ive tried to do this question multiple times but i just cant seem to understand it
Answers
The domain of the function which is the entire x values during the strike is
[tex]0\leq x\leq230[/tex]Related Questions
A sprinkler rotates back and forth from point A to point B. The water reaches 8 meters from the base of the sprinkler.What is the length of the arc AB, rounded to the nearest tenth of a meter? Use 3.14 for [tex]\pi[/tex]
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20.9m
1) Since we want to know the length of that arc, and the angle is written in degrees, let's use a formula to find this out:
[tex]\begin{gathered} l=\frac{\alpha}{360}\cdot2\pi R \\ l=\frac{150}{360}\cdot2(3.14)\cdot8 \\ l=20.93\approx20.9m \end{gathered}[/tex]2) Rounding off to the nearest tenth we have the length of this arc is 20.9, m
Please answer part a and b questions are in the picture
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Part A
we have that
A(4) ----> looking at the graph
A(4) means----> population in the year 1994
so
A(4)----> less than 2 million
and
B(4) -----> greater than 2 million
therefore
the answer Part a is option B
Part B
there is only one value of t where A(t)=B(t)
the value of t is 6 (the year 1996)
Find the slope of the graph of the function at the given point.
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Consider the following function:
[tex]f(x)=\text{ }\tan(x)\text{ cot\lparen x\rparen}[/tex]First, let's find the derivative of this function. For this, we will apply the product rule for derivatives:
[tex]\frac{df(x)}{dx}=\tan(x)\cdot\frac{d}{dx}\text{ cot\lparen x\rparen + }\frac{d}{dx}\text{ tan\lparen x\rparen }\cdot\text{ cot\lparen x\rparen}[/tex]this is equivalent to:
[tex]\frac{df(x)}{dx}=\tan(x)\cdot(\text{ - csc}^2\text{\lparen x\rparen})\text{+ \lparen sec}^2(x)\text{\rparen}\cdot\text{ cot\lparen x\rparen}[/tex]or
[tex]\frac{df(x)}{dx}=\text{ -}\tan(x)\cdot\text{ csc}^2\text{\lparen x\rparen+ sec}^2(x)\cdot\text{ cot\lparen x\rparen}[/tex]now, this is equivalent to:
[tex]\frac{df(x)}{dx}=\text{ -2 csc \lparen2x\rparen + 2 csc\lparen2x\rparen = 0}[/tex]thus,
[tex]\frac{df(x)}{dx}=0[/tex]Now, to find the slope of the function f(x) at the point (x,y) = (1,1), lug the x-coordinate of the given point into the derivative (this is the slope of the function at the point):
[tex]\frac{df(1)}{dx}=0[/tex]Notice that this slope matches the slope found on the graph of the function f(x), because horizontal lines have a slope 0:
We can conclude that the correct answer is:
Answer:The slope of the graph f(x) at the point (1,1) is
[tex]0[/tex]Show exact steps to solve and show the image!Don't mind the pink writing
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1)To construct the line parallel to given line passing through given point, first take a point on the line.
2)Here in the problem that point is Q.
3)Join PQ.
4)After joining PQ, copy the angle made by PQ by constructing the arc MN with steel point of compass on Q. Keep same disttance and get arc M'N' by keeping steel point on P. Then measure length MN on the angle PQR and cut arc by placing steel point on M' and cutting the arc to get point N'.
5) Join PN' and extend till point S.
6) PS is parallel to QR.
a number, twice that number, and one-third of that number added. the result is 20. what is the number?
Answers
Answer:
6
Step-by-step explanation:
Let x = the number
2x = twice the number
1/3 x = one-third of the number
x + 2x + 1/3 x = 20
Combine like terms.
3 1/3 x = 20
Change 3 1/3 to an improper number.
10/3 x = 20
Times by 3/10 on both sides.
3/10 • 10/3 x = 20•3/10
x = 60/10
x = 6
Check:
6 + 2(6) + 1/3(6)
= 6 + 12 + 2
= 20 check!
x+2x+(1/3)x=20...multiply terms by 3
3x+6x+x=60
10x=60
x=6
I need help with this page pls help me !!
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N 6
we have
[tex]216=\frac{r}{2}+214[/tex]a ------> subtraction
subtract 214 both sides
[tex]\begin{gathered} 216-214=\frac{r}{2} \\ 2=\frac{r}{2} \end{gathered}[/tex]b ------> multiplication
Multiply by 2 both sides
[tex]\begin{gathered} 2\cdot2=2\cdot\frac{r}{2} \\ r=4 \end{gathered}[/tex]c ------> r=4
What feature of Kelth's graph makes it difficult to visually compare the responses of those with some college to those shown in the other graphs? (Select all that apply.) (A)The donut hole in the graph made by Keith is a different size than in the graphs made by Ramon. (B)The graphs do not have data labels showing the percentages.(C)The graphs made by Keith and Ramon are all donut pie charts. (D)The graphs made by Keith and Ramon compare groups across education level. (E)The graphs made by Keith and Ramon use the same colors for each of the corresponding responses. 2 How would you change Keith's graph for easier comparison? (Select all that apply.) (A) Make all donuts exactly the same size, with the radius of the holes the same as well. (B)Change the graphs from donut ple charts to time series graphs. (C)Use different sets of colors in each of the donut ple charts.(D)Combine the graphs into one donut pie chart.(E) Add data labels showing the percentages,
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Answer:
(A)The donut hole in the graph made by Keith is a different size than in the graphs made by Ramon.
Explanations:
Ex
Considering the graph made by Keith and those made by Ramon, we would observe that Keith's graph ( Some college) has a smaller donut hole that Ramons's graphs ( High school or less, and college graduate). This difference in the donut holes will make any comparism made between the " some college" graph and other graphs to be inaccurate.
Transforming the graph of a function by reflecting over an axis
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ANSWER:
(a)
(b)
STEP-BY-STEP EXPLANATION:
(a)
We must do the following transformation:
[tex]y=f(x)\rightarrow y=f(-x)[/tex]In this case, reflects f(x) about the y-axis. The rule that follows the above, is like this:
[tex](x,y)\rightarrow(-x,y)[/tex]We apply the rule to the points of the function and it would be:
[tex]\begin{gathered} \mleft(-4.2\mright)\rightarrow(4,2) \\ (0,4)\rightarrow(0,4) \\ (4,6)\rightarrow(-4,6) \end{gathered}[/tex]We graph and we have:
(b)
We must do the following transformation:
[tex]y=g(x)\rightarrow y=-g(x)[/tex]In this case, reflects f(x) about the x-axis. The rule that follows the above, is like this:
[tex](x,y)\rightarrow(x,-y)[/tex]We apply the rule to the points of the function and it would be:
[tex]\begin{gathered} \mleft(-7,-2\mright)\rightarrow\mleft(-7,2\mright) \\ \mleft(-4,-5\mright?)\rightarrow\mleft(-4,5\mright) \\ \mleft(4,-1\mright)\rightarrow\mleft(4,1\mright) \end{gathered}[/tex]We graph and we have:
using the converse of the same-side interior angles postulate what equation shows that g∥h
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Answer: [tex]\angle 2+\angle 4=180^{\circ}[/tex] or [tex]\angle 1+\angle 3=180^{\circ}[/tex]
How many apple pies did they sell and how many blueberry pies did they sell?
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Let the number of apple pies x
Let the number of blue pies y
Since they sold 38 pies on Saturday, then
Add x and y, then equate the sum by 38
[tex]x+y=38\rightarrow(1)[/tex]Since they sold each apple pie for $11 and each blueberry pie for $13
Since they collected $460 on Saturday, then
Multiply x by 11 and y by 13, then add the products and equate the sum by 460
[tex]11x+13y=460\rightarrow(2)[/tex]Now, we have a system of equations to solve it
Multiply equation (1) by -13 to equate the coefficients of y in values and opposite them in signs to eliminate them
[tex]\begin{gathered} (-13)(x)+(-13)(y)=(-13)(38) \\ -13x-13y=-494\rightarrow(3) \end{gathered}[/tex]Add equations (2) and (3)
[tex]\begin{gathered} (11x-13x)+(13y-13y)=(460-494) \\ -2x+0=-34 \\ -2x=-34 \end{gathered}[/tex]Divide both sides by -2
[tex]\begin{gathered} \frac{-2x}{-2}=\frac{-34}{-2} \\ x=17 \end{gathered}[/tex]Substitute the value of x in equation (1) to find y
[tex]17+y=38[/tex]Subtract 17 from both sides
[tex]\begin{gathered} 17-17+y=38-17 \\ y=21 \end{gathered}[/tex]The y sold 17 apple pies and 21 blueberry pies
The answer is the last choice
The table shows the earnings and the number of hours worked for five employees. complete the table by finding the missing values.
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The first employee
[tex]\begin{gathered} He\text{ earns a total of \$12.75} \\ \text{His working rate is \$}8.50\text{ per hour} \\ \text{Hours he workd can be calculated below} \\ \text{ \$8.50 = 1 hour} \\ \text{ \$12.75 =?} \\ \text{ number of hours=}\frac{12.75}{8.50} \\ \text{ number of hours = 1.5 hours} \end{gathered}[/tex]The second employee
[tex]\text{ earning per hour = }\frac{19.09}{2.3}=\text{ \$8.3 per hour}[/tex]The third employee
[tex]\begin{gathered} \text{ \$7.75=1 hour} \\ \text{ \$26.}35=\text{?} \\ \text{ number of hours=}\frac{26.35}{7.75}=3.4\text{ hours} \end{gathered}[/tex]The fourth employee
[tex]\text{earning per hour = }\frac{49.50}{4.5}=\text{ \$}11\text{ per hour}[/tex]The fifth employee
[tex]\text{earning per hour=}\frac{31.50}{1.5}=\text{ \$21 per hour}[/tex]Find the probability of at least 2 girls in 6 births. Assume that male and female births are equally likely and that the births are independent events.0.6560.1090.2340.891
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We need to use Binomial Probability.
Of 6 births, we want to find the probability of at least 2 of them being girls.
To solve this, we need to find:
Probability of exactly 2 girls
Probability of exactly 3 girls
Probability of exactly 4 girls
Probability of exactly 5 girls
Probability of exactly 6 girls
If we add all these probabilities, we get the probability of at least 2 girls.
To find the probabilities, we can use the formula:
[tex]_nC_r\cdot p^r(1-p)^{n-r}[/tex]Where:
n is the number of trials (in this case, the number of total births)
r is the number of girls we want to find the probability
p is the probability of the event occurring
[tex]_nC_r\text{ }is\text{ }the\text{ }combinatoric\text{ }"n\text{ }choose\text{ }r"[/tex]The formula for "n choose r" is:
[tex]_nC_r=\frac{n!}{r!(n-r)!}[/tex]Then, let's find the probability of exactly 2 girls:
The probability of the event occurring is:
[tex]P(girl)=\frac{1}{2}[/tex]Because there is a 50% probability of being a girl or a boy.
let's find "6 choose 2":
[tex]_6C_2=\frac{6!}{2!(6-2)!}=\frac{720}{2\cdot24}=15[/tex]Now we can find the probability of exactly 2 girls:
[tex]Exactly\text{ }2\text{ }girls=15\cdot(\frac{1}{2})^2(1-\frac{1}{2})^{6-2}=15\cdot\frac{1}{4}\cdot(\frac{1}{2})^4=\frac{15}{4}\cdot\frac{1}{16}=\frac{15}{64}[/tex]We need to repeat these calculations for exactly 3, 4, 5, and 6 girls:
Exactly 3 girls:
let's find "6 choose 3":
[tex]_6C_3=\frac{6!}{3!(6-3)!}=\frac{720}{6\cdot6}=20[/tex]Thus:
[tex]Exactly\text{ }3\text{ }girls=20\cdot(\frac{1}{2})^3(1-\frac{1}{2})^{6-3}=20\cdot\frac{1}{8}\cdot\frac{1}{8}=\frac{5}{16}[/tex]Exactly 4 girls:
"6 choose 4":
[tex]_6C_4=\frac{6!}{4!(6-4)!}=\frac{720}{24\cdot2}=15[/tex]Thus:
[tex]Exactly\text{ }4\text{ }girls=15\cdot(\frac{1}{2})^4(1-\frac{1}{2})^{6-4}=15\cdot\frac{1}{16}\cdot\frac{1}{4}=\frac{15}{64}[/tex]Exactly 5 girls:
"6 choose 5"
[tex]_6C_5=\frac{6!}{5!(6-5)!}=\frac{720}{120}=6[/tex]Thus:
[tex]Exactly\text{ }5\text{ }girls=6\cdot(\frac{1}{2})^5(1-\frac{1}{2})^{6-5}=6\cdot\frac{1}{32}\cdot\frac{1}{2}=\frac{3}{32}[/tex]Exactly 6 girls:
"6 choose 6"
[tex]_6C_6=\frac{6!}{6!(6-6)!}=\frac{720}{720\cdot0!}=\frac{720}{720}=1[/tex]Thus:
[tex]Exactly\text{ }6\text{ }girls=1\cdot(\frac{1}{2})^6(1-\frac{1}{2})^{6-6}=\frac{1}{64}\cdot(\frac{1}{2})^0=\frac{1}{64}[/tex]now, to find the answer we need to add these 5 values:
[tex]\frac{15}{64}+\frac{5}{16}+\frac{15}{64}+\frac{3}{32}+\frac{1}{64}=\frac{57}{64}=0.890625[/tex]To the nearest tenth, the probability of at least 3 girls is 0.891, thus, the last option is the correct one.
If 10 g of a radioactive substance are present initially and 9 yr later only 5 g remain, how much of the substance will be present after 18 yr?After 18 yr there will be g of a radioactive substance.(Round the final answer to three decimal places as needed. Round all intermediate values to seven decimal places as needed.)
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Given:
The initial amount of substance, No=10 g.
The amount of substance left after 9 years, N=5 g.
Since 10 g of substance is present initially, and it became 5 g(half of the initial amount) in 9 years, the half life of the substance is, t =9 years.
Hence, the expression for the amount remaining after T years is,
[tex]N(t)=N_0(\frac{1}{2})^{\frac{T}{t_{}}}[/tex]To find the amount of substance remaining after 18 years, put T=18, N0=10 and t=9 in the above equation.
[tex]\begin{gathered} N(18)=10\times(\frac{1}{2})^{\frac{18}{9}} \\ N(18)=10(\frac{1}{2})^2 \\ =\frac{10}{4} \\ =2.5\text{ g} \end{gathered}[/tex]Therefore, after 18 years 2.5 g of the radioactive substance will remain.
hardest question on brainly stumbles college students!
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Applying the vertical angles theorem and other properties, the value of x is 20. The reason for each statement has been explained below.
What are Vertical Angles?A pair of vertical angles are formed when two straight lines intersect each other at a common point. The angles that face each other directly are vertical angles and they are congruent or equal to each other.
Given the diagram below, to find the value of x, the following are the each reason that justifies each of the statements in each step:
Statement Reasons
1. m∠DOB = m∠DOE + m∠BOE 1. Angle Addition Postulate
2. m∠DOB = 90 + x 2. Substitution
3. m∠AOC = m∠DOB 3. Vertical Angels Theorem
4. 110 = 90 + x 4. Substitution
5. x = 20 5. Algebra
To solve using algebra, step 4 is solved as explained below:
110 - 90 = 90 + x - 90
20 = x
x = 20.
Therefore, applying the above steps and reasons that includes the use of vertical angles theorem, the value of x is determined to be: x = 20.
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What is the distance between A(5,-2) and B(-2,4)?
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Answer:
[tex]\sqrt{85}[/tex]
Step-by-step explanation:
Let's use the distance formula to solve for the distance between the two given points!
d = [tex]\sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2 }[/tex]
Now, we input the points:
(5-(-2) + (-2-4)
(which will equal...)
(7) + (-6)
Now we input the solutions we got here to the distance formula:
[tex]d =\sqrt{(7)^2 + (-6)^2[/tex]
(we simplify....)
[tex]7^2 = 49\\(-6)^2 = 36[/tex]
input these solutions into the distance formula again...
[tex]\sqrt{49 + 36} = \sqrt{85}[/tex]
85 is not a number that can be square rooted properly, nor does it have any perfect squares available to divide equally.
Therefore, we conclude that the distance between A(5, -2) and B(-2,4) is [tex]\sqrt{85}[/tex].
Ana has $75 and saves an additional $13 per week. Which equation can be used to findhow many weeks it will take until she has $452?75 + w = 4520 75 + 13w = 45213w = 75 = 452452 + 13w = 75
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Here, we want to get an equation
Firstly, since we do not have the number of weeks, we can represent it with a variable (a letter)
In this case, we shall be representing it with w
Since she saves $13 in a week, in w weeks, the amount saved will be;
13 * w = $13w
Now, recall that she has $75 before she started saving. What this mean is that at the end of the w weeks, the amount she will have will be ;
[tex]13w\text{ + 75}[/tex]We now proceed to equate this to the total she wants to save and we finally have the complete equation below;
[tex]13w\text{ + 75 = 452}[/tex]What value n makes the eauqation n x 3/4 = 3/16
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Answer:
N = 1/4
Step-by-step explanation:
Okay, so 1/4 is equal to N.
3/4 x1/4=3/16
What is 4 1/10 equal to
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We are given the following mixed fraction:
[tex]4\frac{1}{10}[/tex]This is a fraction of the form:
[tex]a\frac{b}{c}[/tex]Any mixed fraction can be rewritten using the following relationship:
[tex]a\frac{b}{c}=a+\frac{b}{c}[/tex]Applying the relationship we get:
[tex]4\frac{1}{10}=4+\frac{1}{10}[/tex]Now, we add the whole number and the fraction using the following relationship:
[tex]a+\frac{b}{c}=\frac{ac+b}{c}[/tex]Applying the relationship we get:
[tex]4+\frac{1}{10}=\frac{40+1}{10}=\frac{41}{10}=4.1[/tex]Therefore, the mixed fraction is equivalent to 4.1
in order to clean her aquarium Stephanie much remove half of the water the garden measures 30 inches long 16 inches wide and 12 inches deep the aquarium is currently completely full with volume of water in cubic inches must Stephanie remove?
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Hello!
30in long (length)
16in wide (width)
12in deep (height)
First, we have to calculate the volume when the aquarium is full of water, using the formula:
[tex]undefined[/tex]what's the leading term and constant of -.5x^5+1.5
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We have the following polynomial:
[tex]-0.5x^5+1.5[/tex]And we have to determine which is the leading term, and the constant term of that polynomial.
1. To determine that we know that the leading term is that term in the polynomial that contains the highest power of the variable. In this case, the variable is x, and the term with the highest variable is:
[tex]-0.5x^5\rightarrow\text{ This is the leading term}[/tex]2. To determine the constant term, we have to remember that this term is not associated with the variable, that is, is not a coefficient of the variable. Therefore, the constant term is 1.5.
Hence, in summary, we have that:
[tex]\text{ Leading term: }-0.5x^5[/tex]And
[tex]\text{ Constant term: }1.5[/tex]Lana draws ALMN on the coordinate plane. What is the perimeter of ALMN? Round to the nearest unit
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We are asked to determine the perimeter of triangle LMN. To do that we will use the fact that the perimeter is the sum of the length of the sides of the triangle. Therefore, we have:
[tex]P=LM+MN+LN[/tex]To determine the value of the length of "LM" we will use the formula for the euclidian distance:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]Where:
[tex]\begin{gathered} (x_1,y_1)_;\left(x_2,y_2\right) \\ \end{gathered}[/tex]Are the endpoints of the segment. For LM we have that the coordinates of the endpoints are:
[tex]L=\lparen-3,2)[/tex][tex]M=(3,5)[/tex]Substituting we get:
[tex]d_{LM}=\sqrt{(3-(-3))^2+(5-2)^2}[/tex]Solving the operations:
[tex]d_{LM}=\sqrt{6^2+3^2}[/tex]Solving the operations:
[tex]d_{LM}=\sqrt{45}[/tex]Now, we use the endpoints of MN:
[tex]M=(3,5)[/tex][tex]N=(9,2)[/tex]Substituting we get:
[tex]d_{MN}=\sqrt{(9-3)^2+(2-5)^2}[/tex]Solving the operations we get:
[tex]\begin{gathered} d_{MN}=\sqrt{6^2+\left(-3\right)^2} \\ \\ d_{MN}=\sqrt{45} \end{gathered}[/tex]Now, we apply the equation for segment LN:
[tex]d_{LN}=\sqrt{}(9-(-3))^2+(2-2)^2[/tex]Solving the operations:
[tex]d_{LN}=12[/tex]Now, we substitute in the formula for the perimeter:
[tex]P=\sqrt{45}+\sqrt{45}+12[/tex]Adding like terms:
[tex]P=2\sqrt{45}+12[/tex]In decimal form rounded to the nearest unit this is:
[tex]P=25[/tex]Therefore, the perimeter of the figure is 25.
HELP PLEASEEEEE!!!!!!
Answers
The one rational number between -0.45 and -0.46 is -0.455.
What is defined as the term rational number?Rational are numbers which can be specified in the form p/q, in which p and q are integers and q≠. The distinction among rational numbers as well as fractions is that fractions cannot include a negative denominator or numerator. As a result, the denominator and numerator of such a fraction have been whole numbers (denominator 0), whereas the numerator and the denominator of rational numbers are integers.For the given question.
The two rational number are given as;
-0.45 and -0.46.
To find the on rational number lying between the two given rational number is to take the average of both numbers.
= (-0.45 + (-0.46))/2
= -0.91/2
= -0.455
Thus, the one rational number between -0.45 and -0.46 is -0.455.
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Classifying systems of linear equations from graphsFor each system of linear equations shown below, classify the system as "consistent dependent," "consistent independent," or "inconsistent." Then, choose thebest description of its solution. If the system has exactly one solution, give its solution.System ASystem B System C
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Consistent dependant system- System B: It has infinite number of solutions, in this case, the graphs of the lines are the same.
Consistent independent system- System C: It has exactly one solution. In this case, both lines cross each other at exactly one point.
Solution : (-2,-2)
Inconsistent: System A.
When a system has no solution, lines never cross each other.
Reece increases the amount of money he pays into his savings account by 4% each year. This year, he paid £3000 into his account. To the nearest penny, how much did Reece pay into his account a) 1 year ago? b) 10 years ago?
Answers
The money deposited 1 year ago is $2884.61 and the money deposited 10 years ago is $2142.85.
Given that, Reece increases the amount of money he pays into his savings account by 4% each year.
What is savings account?A savings account is a bank account at a retail bank. Common features include a limited number of withdrawals, a lack of cheque and linked debit card facilities, limited transfer options and the inability to be overdrawn.
We know that, simple interest = (P×R×T)/100
a) P=$x, R=4% and T=1 year
SI=3000-x
⇒ 3000-x = (x×4×1)/100
⇒ 3000-x=0.04x
⇒ 1.04x=3000
⇒ x=3000/1.04
⇒ x=$2884.61
Money deposited 1 year ago is $2884.61.
b) P=$y, R=4% and T=10 year
SI=3000-y
⇒ 3000-y = (y×4×10)/100
⇒ 3000-y = 0.4y
⇒ 1.4y = 3000
⇒ y=3000/1.4
⇒ y=$2142.85
Therefore, the money deposited 1 year ago is $2884.61 and the money deposited 10 years ago is $2142.85.
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A = P + PRT/100Make P the subject from the formula.
Answers
ANSWER
[tex]P=\frac{100A}{100+RT}[/tex]EXPLANATION
We want to make the subject of the formula in the given equation:
[tex]A=P+\frac{PRT}{100}[/tex]First, factorize the right-hand side of the equation:
[tex]A=P(1+\frac{RT}{100})[/tex]Simplify the bracket:
[tex]A=P(\frac{100+RT}{100})[/tex]Now, divide both sides by the term in the bracket:
[tex]\begin{gathered} \Rightarrow P=A\cdot\frac{100}{100+RT} \\ \Rightarrow P=\frac{100A}{100+RT} \end{gathered}[/tex]That is the answer.
=
A ball is thrown from a height of 156 feet with an initial downward velocity of 8 ft/s. The ball's height h (in feet) after t seconds is given by the following.
h=156-81-161²
How long after the ball is thrown does it hit the ground?
Round your answer(s) to the nearest hundredth.
Answers
The time taken by the ball to hit the ground is 2.88 sec.
What is termed as the distance?Distance is defined as an object's total movement without regard for direction. Distance can be defined as how much surface an object has covered regardless of its starting or closing point.For the given question,
The total height from which the ball is thrown is 156 feet.
Let 'h' be the height after the time 't' sec.
The equation for the relation of the height and the times is;
h = 156 - 8t - 16t²
The initial velocity of the ball is 8 ft/s. .
When the ball hit the ground the height will become zero.
156 - 8t - 16t² = 0.
Divide the equation by -4.
4t² + 2t - 34 = 0
Solve the quadratic equation using the quadratic formula to find the time.
t = 2.88 sec.
Thus, the time taken by the ball to hit the ground is 2.88 sec.
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The correct question is-
A ball is thrown from a height of 156 feet with an initial downward velocity of 8 ft/s . The ball's height h (in feet) after t seconds is given by the following. h=156-8t-16t²
How long after the ball is thrown does it hit the ground?
Round your answer(s) to the nearest hundredth.
Identify the range of the function shown in the graph. 10 8 4 -10-8-4-2 8 10 O A. -2< y < 2 O B. {-2, 2) O C. y is all real numbers OD. Y > 0
Answers
Answer
Option B is correct.
Range: y is all real numbers.
Explanation
The range of a function refers to the region of values where the function can exist. It refers to the values that the dependent variable [y or f(x)] can take on. It is the region around the y-axis that the graph of the function spans.
For this question, we can see that the graph spans over the entire y-axis.
Hence, the range of this function shown in the graph is all real number.
Hope this Helps!!!
Please help with this problem my son is having problems showing his work an understanding how. Solve x2 – 6x = 16 using the quadratic formula method. Show your work. Then describe the solution.
Answers
Solution
We are given the quadratic equation
[tex]x^2-6x=16[/tex]We want to solve by using the quadratic formula method
Note: Given a quadratic equation
[tex]ax^2+bx+c=0[/tex]The formula method is given
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]From
[tex]\begin{gathered} x^2-6x=16 \\ x^2-6x-16=0 \\ \text{Comparing with the general form of a quadratic equation} \\ a=1 \\ b=-6 \\ c=-16 \end{gathered}[/tex]Substituting the parameters intot the quadratic formula
and
Therefore,
[tex]x=8,-2[/tex]At a coffee shop, there is a pot that has a volume of 5.4 L. Find how many cubic centimeters of coffee will completely fill the pot
Answers
Given:
Total volume = 5.4 L.
We know that 1 L is equivalent to 1000 cubic centimetres; hence:
[tex]5.4L\times\frac{1000cm^3}{1L}[/tex]ANSWER
5400 cm³ of coffee will completely fill the pot
Area of a sector A sector with a radius of \maroonD{8\,\text{cm}}8cmstart color #ca337c, 8, start text, c, m, end text, end color #ca337c has an area of \goldE{56\pi\,\text{cm}^2}56πcm
Answers
To find the angle of the sector, follow the steps below.
Step 01: Find the total area of the circle.
The area (A) of a circle with radius r is:
[tex]A=\pi r^2[/tex]Knowing that r = 8 cm, then the area is:
[tex]\begin{gathered} A=8^2\pi \\ A=64\pi\text{ cm}^2 \end{gathered}[/tex]Step 02: Find the central angle.
To find the angle, use proportions.
Knowing that:
When angle = 2π, A = 64π,
Then when angle is x, A = 56π
[tex]\begin{gathered} \frac{x}{2\pi}=\frac{56\pi}{64\pi} \\ \\ \text{ Multiplying both sides by 2}\pi: \\ \frac{x}{2\pi}*2\pi=\frac{56\pi}{64\pi}*2\pi \\ x=\frac{56*2}{64}\pi \\ x=\frac{112}{64}\pi \\ \\ \text{ Dividing both the numerator and the denominator by 16:} \\ x=\frac{\frac{112}{16}}{\frac{64}{16}}\pi \\ x=\frac{7\pi}{4} \end{gathered}[/tex]Answer: The central angle measure is:
[tex]\frac{7\pi}{4}[/tex]