1) Solving that absolute value equation:
|2x+5|=13 Applying the absolute value eq. property
2x +5 = 13 subtracting 5 from both sides
2x = 13-5
2x= 8 Dividing by 2
x =4
2x +5=-13 subtracting 5 from both sides
2x = -13-5
2x = -18 Dividing by 2
x= -9
Then x=4 or x =-9
2) The equations 2x +5 =13 and 2x +15= -13
Resulting in x=4 or x =-9
Prof. Glatt likes 2% milk (2% fat) for her cereal in the morning. Her parents only buy wholemilk (3.5% fat) and non-fat milk (0% fat). While she is visiting her parents, how much of eachtype of milk does she need to mix to get 3 cups of 2% milk. The answer can be rounded to thenearest tenth.linear systems solving algebraically
It is given that there are two types of milk.
One is 3.5% and one is 0%.
Let the number of cups of 3.5% milk be x and the number of cups of 0% milk used be y.
The total should be 3 cups so it follows:
[tex]x+y=3\ldots(i)[/tex]It is also known that the resulting milk is 2% so it follows:
[tex]\begin{gathered} \frac{3.5}{100}x+\frac{0}{100}y=\frac{2}{100}(x+y) \\ \frac{3.5}{100}x=\frac{2}{100}(x+y) \end{gathered}[/tex]Multiply by 100 on both sides to get:
[tex]\begin{gathered} 3.5x=2(x+y) \\ 3.5x=2x+2y \\ 1.5x=2y \\ x=\frac{2}{1.5}y \\ x=\frac{2\times2}{1.5\times2}y \\ x=\frac{4}{3}y \end{gathered}[/tex]Substitute the value of (ii) in (i) to get:
[tex]\begin{gathered} x+y=3 \\ \frac{4}{3}y+y=3 \\ \frac{4+3}{3}y=3 \\ \frac{7}{3}y=3 \\ \frac{3}{7}\times\frac{7}{3}y=\frac{3}{7}\times3 \\ y=\frac{9}{7} \end{gathered}[/tex]Hence the quantity of 0% milk is 9/7 cups.
The quantity of 3.5% milk is given by:
[tex]\begin{gathered} x=\frac{4}{3}y \\ x=\frac{4}{3}\times\frac{9}{7} \\ x=\frac{12}{7} \end{gathered}[/tex]Hence the quantity of 3.5% milk is 12/7 cups.
Jenny borrowed $8000 at a rate of 9%, compounded semiannually. Assuming she makes no payments, how much will she owe after 10Do not round any intermediate computations, and round your answer to the nearest cent.
SOLUTION
We will use the formula
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ where\text{ } \\ A=amount\text{ after 10 years = ?} \\ P=money\text{ borrowed = \$8,000} \\ r=9\%=\frac{9}{100}=0.09 \\ t=time\text{ in years = 10} \\ n=compounding=semi-annualy=2 \end{gathered}[/tex]plugging in, we have
[tex]\begin{gathered} A=8,000(1+\frac{0.09}{2})^{2\times10} \\ A=8,000(1.045)^{20} \\ A=8,000\times2.4117140 \\ A=19,293.7121986 \end{gathered}[/tex]Hence the answer is $19,293.71
Alexa claims that the product of 2.3 and 10^2 is 0.23. Do you agree or disagree? Explain why or why not?
Answer:
disagree
Step-by-step explanation:
product = 2.3 * 10²
= 2.3 * 100
= 230
thus, the answer is different from the one acclaimed by Alexa.
can you solve for x and y y=4x-11=x+13
x = 8, y = 21
Explanations:The given equation is:
y = 4x - 11 = x + 13
This can be splitted into two equations as:
y = 4x - 11..........(1)
y = x + 13..........(2)
Substitute equation (1) into equation (2)
4x - 11 = x + 13
4x - x = 13 + 11
3x = 24
x = 24/3
x = 8
Substitute the value of x into equation (1)
y = 4x - 11
y = 4(8) - 11
y = 32 - 11
y = 21
x = 8, y = 21
hello i am haveing some trouble with ineqalles and can you help with this create a word problem that leads to an inequality by filling in the blanks with your corresponding answer.Twenty subtracted from the product of seven and a number exceeds one hundred.
Step 1
To change word problem to an inequality, you must take the word problem step by step and translate it into an inequality.
Step 2
Take the word problem step by step
[tex]\begin{gathered} \text{Twenty refers to the number }^{\prime}20^{\prime} \\ \text{Twenty subtracted from means 20 was removed from something.} \\ i.e\text{ say x-20} \end{gathered}[/tex][tex]\begin{gathered} \text{Twenty subtract}ed\text{ from the product of seven and a number } \\ \text{Product of seven and a number first} \\ Let\text{ the number be x} \\ so\text{ it now} \\ \text{Product of seven and x=7x } \end{gathered}[/tex][tex]\text{Twenty subtract}ed\text{ from the product of seven and a number}=7x-20[/tex]The distance from the ground of a person riding on a Ferris wheel can be modeled by the equation d equals 20 times the sine of the quantity pi over 30 times t end quantity plus 10 comma where d represents the distance, in feet, of the person above the ground after t seconds. How long will it take for the Ferris wheel to make one revolution?
We have the function d, representing the distance from the ground of a person riding on a Ferris wheel:
[tex]d(t)=20\sin (\frac{\pi}{30}t)+10[/tex]If we consider the position of the person at t = 0, which is:
[tex]d(0)=20\sin (\frac{\pi}{30}\cdot0)+10=20\cdot0+10=10[/tex]This position, for t = 0, will be the same position as when the argument of the sine function is equal to 2π, which is equivalent to one cycle of the wheel. Then, we can find the value of t:
[tex]\begin{gathered} \sin (\frac{\pi}{30}t)=\sin (2\pi) \\ \frac{\pi}{30}\cdot t=2\pi \\ t=2\pi\cdot\frac{30}{\pi} \\ t=60 \end{gathered}[/tex]Then, the wheel will repeat its position after t = 60 seconds.
Answer: 60 seconds.
Miguel has 225 base ball cards. He plans to keep 75 cards and give the rest to his friends.Can Miguel give an equal number of cards to 6 friends? Explain.
We know what Miguel has 225 baseball cards, and he plans on keeping 75, while giving them the rest of them to his friends. We want to know if he can give an equal number to 6 of his friends.
For this objective, we know what he will have to give to his friends an amount of:
[tex]undefined[/tex]Multiply. Write your answer in decimal form: (8 x 10^2)(4 x 10^2)
Answer:
32×10⁴
Step-by-step explanation:
open the bracket
8×4×10^(2+2)
32×10⁴
hope it helps
please mark brainliest
consider the function f(x) whose second derivative is f' '(x)=4x+4sin(x). If f(0)=3 and f'(0)=4, what is f(5)?
Problem: consider the function f(x) whose second derivative is f' '(x)=4x+4sin(x). If f(0)=3 and f'(0)=4, what is f(5)?.
Solution:
Let the function f(x) whose second derivative is:
[tex]f^{\prime\prime}(x)\text{ = 4x+4sin(x)}[/tex]Now, the antiderivative (integral) of the above function would be:
EQUATION 1:
[tex]f^{\prime}(x)=\int f^{\prime\prime}(x)\text{ }dx\text{= }2x^2-4\cos (x)\text{ +C1}[/tex]where C1 is a constant because we have an indefinite integral. Now the antiderivative (integral) of the above function f´(x) is:
[tex]f(x)=\int f^{\prime}(x)\text{ }dx\text{=}\int \text{ (}2x^2-4\cos (x)\text{ +C1)}dx\text{ }[/tex]that is:
EQUATION 2:
[tex]f(x)=\text{ }\frac{2x^3}{3}-4\sin (x)+C1x+\text{ C2}[/tex]where C2 is a constant because we have an indefinite integral.
Now using the previous equation, if f(0)= 3 then:
[tex]3=\text{ C2}[/tex]Now, using equation 1 and the fact that f ´(0) = 4, then we have:
[tex]4=f^{\prime}(0)\text{= }^{}-4\text{ +C1}[/tex]That is:
[tex]4=\text{ }^{}-4\text{ +C1}[/tex]Solve for C1:
[tex]8=\text{ }^{}\text{C1}[/tex]Now, replacing the constants C1 and C2 in equation 2, we have an expression for f(x):
[tex]f(x)=\text{ }\frac{2x^3}{3}-4\sin (x)+8x+3[/tex]Then f(5) would be:
[tex]f(5)=\text{ }\frac{2(5)^3}{3}-4\sin (5)+40+3=\text{ }125.98[/tex]
then the correct answer is:
[tex]f(5)=\text{ }125.98[/tex]A hiker on the Appalachian Trail planned to increase the distance covered by 10% each day. After 7 days, the total distance traveled is 75.897 miles.
part A. We are given that a hiker will increase the distance covered by 10% each day. Let "S" be the distance, then on the first day the distance is:
[tex]S_1[/tex]On the second day, we must add 10% of the first day, we get:
[tex]S_1=S_1+\frac{10}{100}S_1[/tex]Simplifying we get:
[tex]S_2=S_1+0.1S_1=1.1S_1[/tex]On the third day, we add 10% of the second day, we get:
[tex]S_3=S_2+0.1S_2=1.1S_2=(1.1)(1.1)S_1=(1.1)^2S_1[/tex]On the fourth day, we add 10% of the third day, we get:
[tex]S_4=S_3+0.1S_3=1.1S_3=(1.1)^3S_1[/tex]If we continue this pattern and we set "n" as the number of days, then a formula for the distance after "n" days is:
[tex]S_n=(1.1)^{n-1}S_1[/tex]Now, we are given that for n = 7 the distance is 75897, therefore, we substitute n = 7 in the formula:
[tex]S_7=(1.1)^{7-1}S_1[/tex]Substituting the value of the distance:
[tex]75897=(1.1)^{7-1}S_1[/tex]Now we can solve for S1, we do that by dividing both sides by 1.1 together with its
exponent:
[tex]\frac{75897}{(1.1)^{7-1}}=S_1[/tex]Now we solve the operations:
[tex]\frac{75897}{(1.1)^6}=S_1[/tex]Solving the operations:
[tex]42842=S_1[/tex]Therefore, the distance the first day was 42842 miles.
part B. The formula for Sn is the given previously but we replace the known value of S1:
[tex]S_n=42842(1.1)^{n-1}[/tex]Part C. To determine the distance after 10 days, we substitute the value n = 10 in the formula, we get:
[tex]S_{10}=42842(1.1)^{10-1}[/tex]Solving the operations we get:
[tex]S_{10}=101019.19[/tex]Therefore, the distance after 10 days is 101019.19 miles.
Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let X be the random variable representing the time it takes her to complete one review. Assume X is normally distributed. Let X be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.
Find the probability that the mean of a month's reviews will take Yoonie from 3.5 to 4.25 hrs.
a. Give the probability statement and the probability. (Enter exact numbers as integers, fractions, or decimals for the probability statement. Round the probability to four decimal places.
Using the normal distribution and the central limit theorem, the probability that the mean of a month's reviews will take Yoonie from 3.5 to 4.25 hrs is:
[tex]P(3.5 \leq \bar{X} \leq 4.25) = 0.7482[/tex]
Normal Probability DistributionThe z-score of a measure X of a variable that has mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by the rule presented as follows:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The z-score measures how many standard deviations the measure X is above or below the mean of the distribution, depending if the z-score is positive or negative.From the z-score table, the p-value associated with the z-score is found, and it represents the percentile of the measure X in the distribution.By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].The mean and the standard deviation of each review are given as follows:
[tex]\mu = 4, \sigma = 1.2[/tex]
For the sampling distribution of sample means of size 16, the standard error is given as follows:
[tex]s = \frac{1.2}{\sqrt{16}} = 0.3[/tex]
The probability that the mean of a month's reviews will take Yoonie from 3.5 to 4.25 hrs is the p-value of Z when X = 4.25 subtracted by the p-value of Z when X = 3.5, hence:
X = 4.25:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (4.25 - 4)/0.3
Z = 0.83.
Z = 0.83 has a p-value of 0.7967.
X = 3.5:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (3.5 - 4)/0.3
Z = -1.67.
Z = -1.67 has a p-value of 0.0475.
Hence the probability is:
0.7967 - 0.0485 = 0.7482.
The statement is:
[tex]P(3.5 \leq \bar{X} \leq 4.25)[/tex]
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Please answer last oneTo graph F using a graphing utility…Either A,B,C, or DLet me know which option
We have to graph the function F(x) defined as:
[tex]F(x)=\frac{x^2-11x-12}{x+6}[/tex]We can graph it as:
To see the complete graph we have to show the horizontal axis from x = -30 to x = 30 and the vertical axis from y = -80 to y = 80.
Answer: Option B
help meeeeeeeeee pleaseee !!!!!
Because x is continuous, we should use interval notation, the domain is:
D: [1, ∞)
How to find the domain?For a function y = f(x), we define the domain as the set of possible inputs of the function (possible values of x).
To identify the domain, we need to look at the horizontal axis. The minimum value is the one we can see in the left side, and the maximum is the one we could see on the right side.
There we can see that the domain starts at x = 1 and extends to the left, so the notation we can use for the domain is:
D: x ≥ 1
We know that the value x =1 belongs because there is a closed dot there.
The correct option is A, because the domain is continuous (as we can see in the graph), we should use interval notation. In this case the domain can be written as:
D: [1, ∞)
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Noah has a coupon for 30% off at his favorite clothing store he uses it to buy hitting and a pair of jeans Noah paid $28 for jeans after using the coupon what is the regular price of the jeans
$28 after 30% off
28 = regular price * (100 - 30)/100
28 = regular price * 70/100
28 = regular price *0.70
regular price = 28/0.70 = 40
Answer:
Regular price = $40
In the lab, Deandre has two solutions that contain alcohol and is mixing them with each other. Solution A is 10% alcohol and Solution B is 60% alcohol. He uses200 milliliters of Solution A. How many milliliters of Solution B does he use, if the resulting mixture is a 40% alcohol solution?
The percentage of alcohol of a solution i is given by the quotient:
[tex]p_i=\frac{v_i}{V_i},_{}[/tex]where v_i is the volume of alcohol in the solution i and V_i is the volume of the solution i.
From the statement of the problem we know that:
1) Solution A has 10% of alcohol, i.e.
[tex]p_A=\frac{v_A_{}}{V_A}=0.1.\Rightarrow v_A=0.1\cdot V_A.[/tex]2) Solution B has 60% of alcohol, i.e.
[tex]p_B=\frac{v_B}{V_B}=0.6\Rightarrow v_B=0.6\cdot V_B.[/tex]3) The volume of solution A is V_A = 200ml.
4) The resulting mixture must have a percentage of 40% of alcohol, so we have that:
[tex]p_M=\frac{v_M}{V_M}=0.4.[/tex]5) The volume of the mixture v_M is equal to the sum of the volumes of alcohol in each solution:
[tex]v_M=v_A+v_{B\text{.}}_{}[/tex]6) The volume of the mixtureVv_M is equal to the sum of the volumes of each solution:
[tex]V_M=V_A+V_B\text{.}[/tex]7) Replacing 5) and 6) in 4) we have:
[tex]\frac{v_A+v_B}{V_A+V_B_{}}=0.4_{}\text{.}[/tex]8) Replacing 1) and 2) in 7) we have:
[tex]\frac{0.1\cdot V_B+0.6\cdot V_B}{V_A+V_B}=0.4_{}\text{.}[/tex]9) Replacing 3) in 8) we have:
[tex]\frac{0.1\cdot200ml_{}+0.6\cdot V_B}{200ml_{}+V_B}=0.4_{}\text{.}[/tex]Now we solve the last equation for V_B:
[tex]\begin{gathered} \frac{0.1\cdot200ml+0.6\cdot V_B}{200ml_{}+V_B}=0.4_{}, \\ \frac{20ml+0.6\cdot V_B}{200ml_{}+V_B}=0.4_{}, \\ 20ml+0.6\cdot V_B=0.4_{}\cdot(200ml+V_B), \\ 20ml+0.6\cdot V_B=80ml+0.4\cdot V_B, \\ 0.6\cdot V_B-0.4\cdot V_B=80ml-20ml, \\ 0.2\cdot V_B=60ml, \\ V_B=\frac{60}{0.2}\cdot ml=300ml. \end{gathered}[/tex]We must use 300ml of Solution B to have a 40% alcohol solution as the resulting mixture.
Answer: 300ml of Solution B.
Identify the domain and range of the relation. Is the relation a function? Why or why not?
{(-3, 1), (0, 2), (1, 5), (2, 4), (2, 1)}
Domain={-3, 0, 1, 2}, Range={1,2,5,4} and the relation is not a function.
What is a function?A relation is a function if it has only one y-value for each x-value.
The given relation is {(-3, 1), (0, 2), (1, 5), (2, 4), (2, 1)}
The domain is the set of all the first numbers of the ordered pairs.
In other words, the domain is all of the x-values.
Domain={-3, 0, 1, 2}
The Range is the set of all the second numbers of the ordered pairs.
In other words, the range is all of the y-values.
Range={1,2,5,4}
The given relation is not a function because there are two values of y for one value of x. It means 4 and 1 are values of 2.
Hence Domain={-3, 0, 1, 2}, Range={1,2,5,4} and the relation is not a function.
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Which is a way to model 9485 with base ten blocks
The most appropriate way to model 9485 with the base ten blocks is
9 - thousand, 4 - hundred, 8- ten, and 5 - once.
In mathematics, it deals with numbers of operations according to the statements. There are four major arithmetic operators, addition, subtraction, multiplication and division,
Base ten block implies the number of blocks according to the digit representation,
For the given question,
The number represents there are 9485 cubes, which can be represented as 9 - thousand, 4 - hundred, 8- ten, and 5 - once.
Thus, the most appropriate way to model 9485 with the base ten blocks is 9 - thousand, 4 - hundred, 8- ten, and 5 - once.
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choose which group of sets the following number belongs to. Be sure to account for ALL sets. 2/7
A. Real numbers, rational numbers
Explanations:Note:
Real numbers are numbers that can be found on the number line. They include all rational and irrational numbers
Natural numbers are counting numbers. They include 0 and all whole numbers (1, 2, 3, ....)
Rational numbers are numbers that can be expressed as fractions of two integers. eg 2/3, 5/4, etc
Irrational numbers are numbers that cannot be expressed a s fractions of two integers. eg √7, π, etc
2/7 is a real number because it can be found on the number line, and is continuous
Also, 2/7 is a rational number because it is expressed as a fraction of two integers (2 and 7)
many solutions can be found for the system of linear equations represented on the graph?A. no solution B. one solution C. two solution D. Infinity many solutions
The lines are not intersecting. The system of linear equations has a solution only if the lines corresponding to the equations intersect.
The general linear equation is,
y=mx+c, where m is the slope.
The slopes of lines m=2.
Since the graphs are parallel or have the same slope and will never intersect, the system of linear equations have no solution.
$75 dinner, 6.25% tax, 18% tip please show work.You have to find the total cost
According the the information given in the exercise, you know that the cost of the dinner was:
[tex]d=_{}$75$[/tex]Where "d" is the cost of the dinner in dollars.
Convert from percentages to decimal numbers by dividing them by 100:
1. 6.25% tax in decimal for:
[tex]\begin{gathered} tax=\frac{6.25}{100} \\ tax=0.0625 \\ \end{gathered}[/tex]2. 18% tip in decimal form:
[tex]\begin{gathered} tip=\frac{18}{100} \\ \\ tip=0.18 \end{gathered}[/tex]To find the amount in dollars of the tax and the the amount in dollars of the tip, multiply "d" by the decimals found above.
Knowing the above, let be "t" the total cost in dollars.
This is:
[tex]\begin{gathered} t=d+0.0625d+0.18d \\ t=75+(0.0625)(75)+(0.18)(75) \\ t=93.1875 \end{gathered}[/tex]Therefore the answer is: The total cost is $93.1875
A cake is cut into 12 equal slices. After 3 days Jake has eaten 5 slices. What is his weekly rate of eating the cake?
5
36
35
36
cakes/week
cakes/week
1
1 cakes/week
35
01. cakes/week
4
Answer:
11.2 Slices / Week
Step-by-step explanation:
We know that Jake has eaten 5 slices of cake in 3 days. You can divide 5 / 3 to get an average of 1.6 slices of cake being eaten per day. The question asks what the weekly rate or eating the cake will be, do you need to multiple 1.6 x 7 for the total amount of cake eaten per week, which is 11.2 slices!
Answer:
11.6
explanation
we have 7 days.
7days-3days =4
in 3 days he has eaten 5 slices
again 4-3 days=1
so in 6 days he has eaten 10 slices
we have 1 day left.so if he eats 5 slices in 3 day,how many he eat slices in 1 day?5/3=1.6
10+1.6=11.6
Drag the tiles to the correct boxes. Not all tiles will be used.
Match each equation with a value of x that satisfies it.
18
1
9
2
5
(x - 2) = 2
√²+7=4
V1-x
= -1
-3
For a given exponential expression, the determined value is x=3,0,6.
What are exponential expressions?A component of an exponential expression is an exponent. Powers can be expressed succinctly using exponential expressions. The exponent represents the number of times the base has been multiplied.Powers can be expressed succinctly using exponential expressions. The exponent represents the number of times the base has been multiplied. Exponential expressions or the representation of multiplication with exponents can be streamlined to produce the most efficient notation possible.Each exponential expression's x value is evaluated.
Therefore,
1. [tex]$ \sqrt{x^2+7}=4 \\[/tex]
[tex]&\left(x^2+7\right)=4^2 \\[/tex]
[tex]&\left(x^2+7\right)=16 \\[/tex]
simplifying the above equation, then we get
x² = 16 - 7 = 9
x = 3
2. [tex]$\sqrt[2]{1-x}=-1$[/tex]
(1 -x) = (-1)²
1 - x = 1
x = 0
3. [tex](x-2)^{\frac{1}{2}}=2 \\[/tex]
(x - 2) = 2²
x - 2 = 4
x = 6
The determined value is x=3,0,6 for a given exponential expression.
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A regular hexagon has sides 2 feet long. What is the exact area of the hexagon? What is the approximate area of the hexagon?
The formula for the area of a hexagon is
[tex]A=\frac{3\sqrt[]{3}}{2}s^2[/tex]where 's' is the length of one side of the regular hexagon.
The side of our regular hexagon is 2 feet, therefore, its area is
[tex]\begin{gathered} A=\frac{3\sqrt[]{3}}{2}\cdot(2)^2=6\sqrt[]{3} \\ 6\sqrt[]{3}=10.3923048454\ldots\approx10 \end{gathered}[/tex]The exact area of the hexagon is 6√3 ft², which is approximately 10 ft².
For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chumpkin hide
Let x and y be the number of holes dug by the chipmunk and the squirrel, respectively.
Therefore, the number of hidden acorns by each animal is given by the equations below
[tex]\begin{gathered} a_{chipmunk}=3x \\ a_{squirrel}=4y \end{gathered}[/tex]On the other hand, since the squirrel needed 4 fewer holes, and the number of hidden acorns is the same
[tex]\begin{gathered} y=x-4 \\ and \\ a_{chipmunk}=a_{squirrel} \end{gathered}[/tex]Thus,
[tex]\begin{gathered} \Rightarrow3x=4y \\ \Rightarrow3x=4(x-4) \\ \Rightarrow3x=4x-16 \\ \Rightarrow x=16 \end{gathered}[/tex]Hence,
[tex]\Rightarrow16*3=48[/tex]The chipmunk hid 48 acorns.Could you solve the table
The relation is decreasing by a factor of 2 each time, so:
[tex]\begin{gathered} y-9=-2(x-0) \\ y=-2x+9 \end{gathered}[/tex]Therefore:
[tex]\begin{gathered} y(100)=-2(100)+9 \\ y(100)=-200+9 \\ y(100)=-191 \end{gathered}[/tex]Answer:
-191
If it costs $1.50 for a pack of Starbursts at ShopRite, how much will 5 packs cost?
Answer : $7.5
1 pack of starbursts cost $1.50 at shoprite.
How much will 5 packs cost
Let the cost in dollars of 5 packs of starbursts be x
1 pack will cost $1.50
5 packs will cost $x
Mathematically,
1 pack ---------------- $1.50
5 packs -------------= $x
Cross multiply
1 * x = 5 x 1.50
x = $7.5
Hence, 5 packs of starbursts would cost $7.5
I have a calculus question about related rates, pic included
ANSWER
40807 cm³/min
EXPLANATION
The tank has the shape of a cone, with a total height of 9 meters and a diameter of 3.5 m - so the radius, which is half the diameter, is 1.75 m. As we can see, the relationship between the height of the cone and the radius is,
[tex]\frac{r}{h}=\frac{1.75m}{9m}=\frac{7}{36}\Rightarrow r=\frac{7}{36}h[/tex]So the volume of water will be given by,
[tex]V(h)=\frac{1}{3}(\pi r^2)h=\frac{1}{3}\cdot\pi\cdot\frac{7^2}{36^2}h^2\cdot h=\frac{49\pi}{3888}h^3[/tex]Where h is the height of the water (not the tank).
If we derive this equation, we will find the rate at which the volume of water is changing with time,
[tex]\frac{dV}{dt}=\frac{49\pi}{3888}\cdot3h^{3-1}=\frac{49\pi}{3888}\cdot3h^2=\frac{49\pi}{1296}h^2[/tex]We want to know what is the change of volume with respect to time, and this is,
[tex]\frac{dV}{dt}=\frac{dV}{dt}\cdot\frac{dh}{dt}[/tex]Because the height also changes with time. We know that this change is 24 cm per minute when the height of the water in the tank is 1 meter (or 100 cm), so we have,
[tex]\frac{dV}{dt}=\frac{49\pi}{1296}h^2\cdot\frac{dh}{dt}=\frac{49\pi}{1296}\cdot100^2cm^2\cdot\frac{24cm}{1min}\approx28507cm^3/min[/tex]This is the rate at which the water is increasing in the tank. However, we know that there is a leak at a rate of 12300 cm³/min, which means that in fact the water is being pumped into the tank at a rate of,
[tex]28507cm^3/min+12300cm^3/min=40807cm^3/min[/tex]Hence, the water is being pumped into the tank at a rate of 40807 cm³/min, rounded to the nearest whole cm³/min.
ok so this is multiplying decimals 7.3 x9.6=please show your work and answer thank you
therefore, the answer is 70.08
Explanation
Step 1
first multiply as if there is no decimal
[tex]\begin{gathered} 7.3\cdot9.6 \\ a)7.3\cdot9.6\Rightarrow73\cdot96 \\ 73\cdot69=7008 \end{gathered}[/tex]Step 2
count the number of digits after the decimal in each factor.
[tex]\begin{gathered} 7.3\Rightarrow1\text{ decimal} \\ 9.6\Rightarrow1\text{ decimal} \\ \text{total }\Rightarrow2\text{ decimals} \end{gathered}[/tex]
Step 3
Put the same number of digits behind the decimal in the product
[tex]7008\Rightarrow put\text{ 2 decimal, }\Rightarrow so\Rightarrow70.08[/tex]therefore, the answer is 70.08
I hope this helps you
What is the volume of a hemisphere with a radius of 6.5 in, rounded to the nearesttenth of a cubic inch?
To calculate the volum of a hemisphere
We use the formula;
V = (2/3)πr³
where r = radius
π is a constant equal 3.14
r= 6.5 in and π = 3.14
Substituting into the formula
V = (2/3) x 3.14 x (6.5)³
Evauluate
V = (2/3) x 3.14 x 274.625
V = (2/3) x 862.3225
V=574.8816666666667
V= 574.89 in³ to the nearest tenth of a cubic inch.
How many roots does x^2-6x+9 have ? It may help to graph the equation.
The roots are those values that make a function or polynomial take a zero value. The roots are also the intersection points with the x-axis. In the case of a quadratic equation you can use the quadratic formula to find its roots:
[tex]\begin{gathered} ax^2+bx+c=y\Rightarrow\text{ Quadratic equation in standard form} \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}\Rightarrow\text{ Quadratic formula} \end{gathered}[/tex]So, in this case, you have
[tex]\begin{gathered} y=x^2-6x+9 \\ a=1 \\ b=-6 \\ c=9 \end{gathered}[/tex][tex]\begin{gathered} x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4(1)(9)}}{2(1)} \\ x=\frac{6\pm\sqrt[]{36-36}}{2} \\ x=\frac{6\pm0}{2} \\ x=\frac{6}{2} \\ x=3 \end{gathered}[/tex]As you can see, this function only has one root, at x = 3.
You can see this in the graph of the function:
