Nora needs to order some new supplies for the restaurant where she works. Therestaurant needs at least 478 forks. There are currently 286 forks. If each set on salecontains 12 forks, write and solve an inequality which can be used to determine s, thenumber of sets of forks Nora could buy for the restaurant to have enough forks.<
Nora needs to order some new supplies for the restaurant where she works. The
restaurant needs at least 478 forks. There are currently 286 forks. If each set on sale
contains 12 forks, write and solve an inequality which can be used to determine s, the
number of sets of forks Nora could buy for the restaurant to have enough forks.
Let
s -----> the number of sets of forks Nora could buy for the restaurant to have enough forks
so
the inequality that represent this situation is
[tex]286+12s\ge478[/tex]solve for s
[tex]\begin{gathered} 12s\ge478-286 \\ 12s\ge192 \\ s\ge16 \end{gathered}[/tex]the minimum number of sets is 16Perform the indicated operation of multiplication or division on the rational expression and simplify
The division of two fractions is the same as multiplying the first by the inverted second fraction:
Then, in this case:
[tex]\frac{24y^2}{5x^2}\div\frac{6y^3}{25x^2}=\frac{24y^2}{5x^2}\times\frac{25x^2}{6y^3}[/tex]Step 2: multiplication of two fractionsWe multiply two fractions by multiplying the numerators and the denominators:
[tex]\frac{24y^2}{5x^2}\times\frac{25x^2}{6y^3}=\frac{24y^2\times25x^2}{5x^2\times6y^3}[/tex]Step 3: simplifying the numbers of the fractionWe know that
[tex]\frac{25}{5}=5\text{ and }\frac{24}{6}=4[/tex]Then, we can use this in our fraction:
[tex]\begin{gathered} \frac{24y^2\times25x^2}{5x^2\times6y^3}=5\cdot4\frac{y^2x^2}{x^2y^3} \\ \downarrow\text{ since 5}\cdot4=20 \\ 5\cdot4\frac{y^2x^2}{x^2y^3}=20\frac{y^2x^2}{x^2y^3} \end{gathered}[/tex]Step 4: exponents of the resultWe know that if we have a division of same base expressions (same letters), the exponent is just a substraction:
[tex]\begin{gathered} \frac{y^2}{y^3}=y^{2-3}=y^{-1} \\ \frac{x^2}{x^2}=x^{2-2}=x^0=1 \end{gathered}[/tex]Then,
[tex]20\frac{y^2x^2}{x^2y^3}=20y^{-1}\cdot1=20y^{-1}[/tex]Since negative exponents correspond to a division, then we can express the answer in two different ways:
[tex]20y^{-1}=\frac{20}{y}[/tex]Answer:[tex]20y^{-1}=\frac{20}{y}[/tex]which graph show the solution set for -1.1×+6.4>-1.3
Problem
-1.1x + 6.4 > - 1.3
Concept
Solve for x by collecting like terms.
Given the figure below, determine the angle that is a same side interior angle with respect to1. To answer this question, click on the appropriate angle.
Same side interior angles are angles on the same side of the transversal line, inside the two lines intersected.
<5 is an interior angle, on the same side as <3.
On The left side of the bisector line.
Which phrase represents the algebraic expression for n-4.A) The quotient of a number and 4. B) 4 less than a number. C) 4 minus a number. D) 4 more than a number.
B) 4 less than a number.
Last year, Bob had $10,000 to invest. He invested some of it in an account that paid 10% simple interest per year, and he invested the rest in an account that paid 8% simple interest per year. After one year, he received a total of $820 in interest. How much did he invest in each account?
Given:
The total amount is P = $10,000.
The rate of interest is r(1) = 10% 0.10.
The other rate of interest is r(2) = 8%=0.08.
The number of years for both accounts is n = 1 year.
The total interest earned is A = $820.
The objective is to find the amount invested in each account.
Explanation:
Consider the amount invested for r(1) as P(1), and the interest earned as A(1).
The equation for the amount obtained for r(1) can be calculated as,
[tex]\begin{gathered} A_1=P_1\times n\times r_1 \\ A_1=P_1\times1\times0.1 \\ A_1=0.1P_1\text{ . . . . .(1)} \end{gathered}[/tex]Consider the amount invested for r(2) as P(2), and the interest earned as A(2).
The equation for the amount obtained for r(2) can be calculated as,
[tex]\begin{gathered} A_2=P_2\times n\times r_2 \\ A_2=P_2\times1\times0.08 \\ A_2=0.08P_2\text{ . . . . . (2)} \end{gathered}[/tex]Since, it is given that the total interest earned is A=$820. Then, it can be represented as,
[tex]A=A_1+A_2\text{ . . . . . (3)}[/tex]On plugging the obtained values in equation (3),
[tex]820=0.1P_1+0.08P_2\text{ . . . . .(4)}[/tex]Also, it is given that the total amount is P = $10,000. Then, it can be represented as,
[tex]\begin{gathered} P=P_1+P_2 \\ 10000=P_1+P_2 \\ P_1=10000-P_2\text{ . }\ldots\ldots.\text{. .(3)} \end{gathered}[/tex]Substitute the equation (3) in equation (4).
[tex]undefined[/tex]Toni decides to plant a 2-foot wide rectangular flower garden along one side of the pool and patio but outside the fence. She measures the length of the fence to be 44 feet long. What is the area of the flower garden?
If she decides to plant a 2-foot wide rectangular flower garden along one side of the pool and patio but outside the fence. She measures the length of the fence to be 44 feet long. The area of the flower garden is 88 square feet.
Area of the flower gardenUsing this formula to determine the area of the flower garden
Area = Width × Length
Where:
Width = 2 feet
Length = 44 feet
Let plug in the formula
Area = 2 × 44
Area = 88 square feet
Therefore the area is 88 square feet.
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The cost to mail a package is 5.00. Noah has postcard stamps that are worth 0.34 and first-class stamps that are worth 0.49 each. An equation that represents this is 0.49f + 0.34p = 5.00Solve for f and p.If Noah puts 7 first-class stamps, how many postcard stamps will he need?
ANSWER
[tex]\begin{gathered} f=\frac{5.00-0.34p}{0.49} \\ p=\frac{5.00-0.49f}{0.34} \\ p=4.618\approx5\text{ postcard stamps} \end{gathered}[/tex]EXPLANATION
The equation that represents the situation is:
[tex]0.49f+0.34p=5.00[/tex]To solve for f, make f the subject of the formula from the equation:
[tex]\begin{gathered} 0.49f=5.00-0.34p \\ \Rightarrow f=\frac{5.00-0.34p}{0.49} \end{gathered}[/tex]To solve for p, make p the subject of the formula from the equation:
[tex]\begin{gathered} 0.34p=5.00-0.49f \\ \Rightarrow p=\frac{5.00-0.49f}{0.34} \end{gathered}[/tex]To find how many postcard stamps Noah will need if he puts 7 first-class stamps, solve for p when f is equal to 7.
That is:
[tex]\begin{gathered} p=\frac{5.00-(0.49\cdot7)}{0.34} \\ p=\frac{5.00-3.43}{0.34}=\frac{1.57}{0.34} \\ p=4.618\approx5\text{ postcard stamps} \end{gathered}[/tex]Find the expression for the possible width of the rectangle.
Given the area of the rectangle is given by the following expression:
[tex]A=x^2+5x+6[/tex]The area of the rectangle is the product of the length by the width
So, we will factor the given expression
To factor the expression, we need two numbers the product of them = 6
and the sum of them = 5
So, we will factor the number 6 to find the suitable numbers
6 = 1 x 6 ⇒ 1 + 6 = 7
6 = 2 x 3 ⇒ 2 + 3 = 5
So, the numbers are 2 and 3
The factorization will be as follows:
[tex]A=(x+3)(x+2)[/tex]So, the answer will be the possible dimensions are:
[tex]\begin{gathered} \text{Length}=x+3 \\ \text{Width}=x+2 \end{gathered}[/tex]Suppose that 27 percent of American households still have a traditional phone landline. In a sample of thirteen households, find the probability that: (a)No families have a phone landline. (Round your answer to 4 decimal places.) (b)At least one family has a phone landline. (Round your answer to 4 decimal places.) (c)At least eight families have a phone landline.
Answer:
(a) P = 0.0167
(b) P = 0.9833
(c) P = 0.0093
Explanation:
To answer these questions, we will use the binomial distribution because we have n identical events (13 households) with a probability p of success (27% still have a traditional phone landline). So, the probability that x families has a traditional phone landline can be calculated as
[tex]\begin{gathered} P(x)=nCx\cdot p^x\cdot(1-p)^x \\ \\ \text{ Where nCx = }\frac{n!}{x!(n-x)!} \end{gathered}[/tex]Replacing n = 13 and p = 27% = 0.27, we get:
[tex]P(x)=13Cx\cdot0.27^x\cdot(1-0.27)^x[/tex]Part (a)
Then, the probability that no families have a phone landline can be calculated by replacing x = 0, so
[tex]P(0)=13C0\cdot0.27^0\cdot(1-0.27)^{13-0}=0.0167[/tex]Part (b)
The probability that at least one family has a phone landline can be calculated as
[tex]\begin{gathered} P(x\ge1)=1-P(0) \\ P(x\ge1)=1-0.167 \\ P(x\ge1)=0.9833 \end{gathered}[/tex]Part (c)
The probability that at least eight families have a phone landline can be calculated as
[tex]P(x\ge8)=P(8)+P(9)+P(10)+P(11)+P(12)+P(13)[/tex]So, each probability is equal to
[tex]\begin{gathered} P(8)=13C8\cdot0.27^8\cdot(1-0.27)^{13-8}=0.0075 \\ P(9)=13C9\cdot0.27^9\cdot(1-0.27)^{13-9}=0.0015 \\ P(10)=13C10\cdot0.27^{10}\cdot(1-0.27)^{13-10}=0.0002 \\ P(11)=13C11\cdot0.27^{11}\cdot(1-0.27)^{13-11}=0.00002 \\ P(12)=13C12\cdot0.27^{12}\cdot(1-0.27)^{13-12}=0.000001 \\ P(13)=13C13\cdot0.27^{13}\cdot(1-0.27)^{13-13}=0.00000004 \end{gathered}[/tex]Then, the probability is equal to
P(x≥8) = 0.0093
Therefore, the answers are
(a) P = 0.0167
(b) P = 0.9833
(c) P = 0.0093
Use this information to answer the following two questions. Mathew finds the deepest part of the pond to be 185 meters. Mathew wants to find the length of a pond. He picks three points and records the measurements, as shown in the diagram. Which measurement describes the depth of the pond? Hide All Z between 13 and 14 meters 36 m 14 m between 14 and 15 meters between 92 and 93 meters Х ag between 93 and 94 meters
it's letter A. Between 13 and 14 meters
Because one side measure 14, and the height (depth) could not be
higher than 14 meters .
The length of the pond can be calculated using the Pythagorean theorem
length^2 = 36^2 + 14^2
length^2 = 1296 + 196
length^2 = 1492
length = 38.6 m
Write the equation of the function in the graph.. Please show all of your work so i can understand
The vertex form of a parabola is:
[tex]y=a(x-h)^2+k[/tex]where (h, k) is the vertex of the parabola and a is some constant.
From the graph, the vertex is located at (1, 4), that is, h = 1 and k = 4.
Substituting with these values and the point (0, 3), we get:
[tex]\begin{gathered} 3=a(0-1)^2+4 \\ 3-4=a(-1)^2 \\ -1=a\cdot1 \\ -\frac{1}{1}=a \\ -1=a \end{gathered}[/tex]Then, the equation of the function is:
[tex]\begin{gathered} y=-1(x-1)^2+4 \\ y=-(x-1)^2+4 \end{gathered}[/tex]Find the solution of this system of linearequations. Separate the x- and y- values with acomma. Enclose them in a pair of parantheses. System of equations4x + 8y = 838x + 7y = 76- 8x - 16y = -1668x + 7y = 76
Given,
System of equation is,
[tex]\begin{gathered} 4x+8y=83\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots(i) \\ 8x+7y=76\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots(ii) \end{gathered}[/tex]Taking the equation (i) as,
[tex]\begin{gathered} 4x+8y=83 \\ 4x=83-8y \\ x=\frac{83-8y}{4} \end{gathered}[/tex]Substituting the value of x in equation (ii) then,
[tex]\begin{gathered} 8x+7y=76 \\ 8(\frac{83-8y}{4})+7y=76 \\ 664-64y+28y=304 \\ 36y=360 \\ y=10 \end{gathered}[/tex]Substituting the value of y in above equation then,
[tex]\begin{gathered} x=\frac{83-8\times10}{4} \\ x=\frac{3}{4} \end{gathered}[/tex]Hence, the value of x is 3/4 and y is 10. (3/4, 10)
System of equation is,
[tex]\begin{gathered} -8x-16y=-166\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots(i) \\ 8x+7y=76\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots(ii) \end{gathered}[/tex]Taking the equation (i) as,
[tex]\begin{gathered} -8x-16y=-166 \\ 8x+16y=166 \\ 4x+8y=83 \\ 4x=83-8y \\ x=\frac{83-8y}{4} \end{gathered}[/tex]Substituting the value of x in equation (ii) then,
[tex]\begin{gathered} 8x+7y=76 \\ 8(\frac{83-8y}{4})+7y=76 \\ 664-64y+28y=304 \\ 36y=360 \\ y=10 \end{gathered}[/tex]Substituting the value of y in above equation then,
[tex]\begin{gathered} x=\frac{83-8\times10}{4} \\ x=\frac{3}{4} \end{gathered}[/tex]Hence, the value of x is 3/4 and y is 10. (3/4, 10)
Which of the following is the result of using the remainder theorem to find F(-2) for the polynomial function F(x) = -2x³ + x² + 4x-3?
Solution
We have the polynomial
[tex]f(x)=-2x^3+x^2+4x-3[/tex]Usin the remainder theorem, we find f(-2) by substituting x = -2
So we have
[tex]\begin{gathered} f(x)=-2x^{3}+x^{2}+4x-3 \\ \\ f(-2)=-2(-2)^3+(-2)^2+4(-2)-3 \\ \\ f(-2)=-2(-8)+4-8-3 \\ \\ f(-2)=16+4-8-3 \\ \\ f(-2)=20-11 \\ \\ f(-2)=9 \end{gathered}[/tex]Therefore, the remainder is
[tex]9[/tex]I need help finding 5 points. the vertex, 2 to the left of the vertex, and 2 points to the right of the vertex.
Let's convert the given equation first into a vertex form.
[tex]y=a(x-h)^2+k[/tex]where (h, k) is the vertex.
The vertex form of the equatio that we have is:
[tex]y=-2(x-0)^2+0[/tex]Hence, the vertex of the equation is at the origin (0, 0).
Since "a" is negative, our parabola is opening downward.
Let's identify two points to the left of the vertex. Let's say at x = -1. Replace "x" with -1 in the equation.
[tex]\begin{gathered} y=-2(-1)^2 \\ y=-2(1) \\ y=-2 \end{gathered}[/tex]Hence, we have a point to the left of the parabola at (-1, -2).
Let's say x = -2. Replace "x" with -2 in the equation.
[tex]\begin{gathered} y=-2(-2)^2 \\ y=-2(4) \\ y=-8 \end{gathered}[/tex]Hence, we also have another point to the left of the parabola at (-2, -8).
If our x is to the right of the vertex, say, x = 1. Replace "x" with 1 in the equation.
[tex]\begin{gathered} y=-2(1)^2 \\ y=-2(1) \\ y=-2 \end{gathered}[/tex]We have a point to the right of the parabola at (1, -2).
If x = 2, let's replace "x" with 2 in the equation.
[tex]\begin{gathered} y=-2(2)^2 \\ y=-2(4) \\ y=-8 \end{gathered}[/tex]Hence, we also have another point to the right of the parabola at (2, -8).
The graph of this equation is:
Given the formula for the nth term, state the first 5 terms of each sequence.t1= 800, tn= -0.25tn-1
In this case, we'll have to carry out several steps to find the solution.
Step 01:
data:
t1 = 800
tn = - 0.25 tn-1
Step 02:
sequence:
t1 = 800
t2 = -0.25 (800) = - 200
t3 = -0.25 (-200) = 50
t4 = -0.25 (50) = -12.5
t5 = - 0.25 (-12.5) = 3.125
The answer is:
t1 = 800
t2 = - 200
t3 = 50
t4 = -12.5
t5 = 3.125
See photo for problem
Answer:
possible outcome= {H,T}
number of possible outcome=2
obtaining a tail(T)=1
n(T)=1
P(T)=n(T)/number of possible outcome
=1/2
Pls help ASAP!!! Ill give you 5.0
The equivalent equation of 6x + 9 = 12 is 2x + 3 = 4.
Another equivalent equation of 6x + 9 = 12 is 3x + 4.5 = 6
What are equivalent equations?Equivalent equations are algebraic equations that have identical solutions or roots. In other words, equivalent equations are equations that have the same answer or solution.
Therefore, the equivalent equation of 6x + 9 = 12 can be calculated as follows:
6x + 9 = 12
Divide through by 3
6x / 3 + 9 / 3 = 12 / 3
2x + 3 = 4
Therefore, the equivalent equation of 6x + 9 = 12 is 2x + 3 = 4
Another equation that is equivalent to 6x + 9 = 12 is 3x + 4.5 = 6
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hello I'm stuck on this question and need help thank you
Explanation
[tex]\begin{gathered} -2x+3y\ge9 \\ x\ge-5 \\ y<6 \end{gathered}[/tex]Step 1
graph the inequality (1)
a) isolate y
[tex]\begin{gathered} -2x+3y\geqslant9 \\ add\text{ 2x in both sides} \\ -2x+3y+2x\geqslant9+2x \\ 3y\ge9+2x \\ divide\text{ both sides by 3} \\ \frac{3y}{3}\geqslant\frac{9}{3}+\frac{2x}{3} \\ y\ge\frac{2}{3}x+3 \end{gathered}[/tex]b) now, change the symbol to make an equality and find 2 points from the line
[tex]\begin{gathered} y=\frac{2}{3}x+3 \\ i)\text{ for x=0} \\ y=\frac{2}{3}(0)+3 \\ \text{sp P1\lparen0,3\rparen} \\ \text{ii\rparen for x=3} \\ y=\frac{2}{3}(3)+3=5 \\ so\text{ P2\lparen3,5\rparen} \end{gathered}[/tex]now, draw a solid line that passes troguth those point
(0,3) and (3,5)
[tex]y\geqslant\frac{2}{3}x+3\Rightarrow y=\frac{2}{3}x+3\text{\lparen solid line\rparen}[/tex]as we need the values greater or equatl thatn the function, we need to shade the area over the line
Step 2
graph the inequality (2)
[tex]x\ge-5[/tex]this inequality represents the numbers greater or equal than -5 ( for x), so to graph the inequality:
a) draw an vertical line at x=-5, and due to we are looking for the values greater or equal than -5 we need to use a solid line and shade the area to the rigth of the line
Step 3
finally, the inequality 3
[tex]y<6[/tex]this inequality represents all the y values smaller than 6, so we need to draw a horizontal line at y=6 and shade the area below the line
Step 4
finally, the solution is the intersection of the areas
I hope this helps you
what operation helps calculate unit rates and unit prices
Division operation is operation helps calculate unit rates and unit prices.
Division operation -
A rate with 1 as the denominator is referred to as a unit rate. If you have a rate, such as a price per a certain number of items, and the quantity in the denominator is not 1, you can determine the unit rate or price per unit by performing the division operation: numerator divided by denominator.What method do you employ to determine the unit rate?
Simple division of the numerator and denominator yields the unit rate. The outcome tells us how many of the units in the numerator to anticipate for each unit in the denominator.
What in mathematics are rate and unit rate?
A ratio called a rate compares two amounts of DIFFERENT types of UNITS. When expressed as a fraction, a unit rate has a denominator of 1. Divide the rate's numerator and denominator by the denominator to represent the rate as a unit rate.Learn more about Division operation
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A card is drawn from a deck of 52 cards. What is the probability that it is a numbered card (2-10) or a heart?
we know that
Total cards=52
Total numbered card (2-10)=36
Total heart=13
numbered card and heart=9
therefore
The probability is equal to
P=(36+13-9)/52
P=40/52
P=20/26=10/13
The answer is 10/13Suppose that y varies inversely with x, and y = 5/4 when x = 16.(a) Write an inverse variation equation that relates x and y.Equation: (b) Find y when x = 4.y =
In general, an inverse variation relation has the form shown below
[tex]\begin{gathered} y=\frac{k}{x} \\ k\to\text{ constant} \end{gathered}[/tex]It is given that x=16, then y=5/4; thus,
[tex]\begin{gathered} \frac{5}{4}=\frac{k}{16} \\ \Rightarrow k=\frac{5}{4}\cdot16 \\ \Rightarrow k=20 \end{gathered}[/tex]Therefore, the equation is y=20/x
[tex]\Rightarrow y=\frac{20}{x}[/tex]2) Set x=4 in the equation above; then
[tex]\begin{gathered} x=4 \\ \Rightarrow y=\frac{20}{4}=5 \\ \Rightarrow y=5 \end{gathered}[/tex]When x=4, y=5.
im doing math and im wondering when do i switch the inequality?
Question:
Solve the following inequality:
[tex]12x+6<17[/tex]Solution:
Consider the following inequality
[tex]12x+6<17[/tex]solving for 12x, we get:
[tex]12x<17-6[/tex]this is equivalent to:
[tex]12x<11[/tex]solving for x, we get:
[tex]x<\frac{11}{12}[/tex]so that, the correct answer is:
[tex]x<\frac{11}{12}[/tex]a= 8 in, b= ? C= 14 in.using pythagorean theorem
by Pythagorean theorem'
[tex]8^2+b^2=14^2[/tex][tex]\begin{gathered} 64+b^2=196 \\ b^2=196-64 \end{gathered}[/tex][tex]\begin{gathered} b^2=132 \\ b=\sqrt[]{132} \\ b=11.48 \end{gathered}[/tex]b = 11.48
Four more than three times a number, is less than 30. Which of the following is not a solution?61278
Solution
- To solve the question, we simply need to interpret the question line by line.
- Let the number be x.
- "Four more than three times a number" can be written as:
[tex]\begin{gathered} \text{ Three times a number is: }3x \\ \text{ For more than three times a number becomes: }4+3x \end{gathered}[/tex]- "Four more than three times a number is less than 30" can be written as:
[tex]4+3x<30[/tex]- Now, we can proceed to solve the inequality and find the appropriate range of x. This is done below:
[tex]\begin{gathered} 4+3x<30 \\ \text{ Subtract 4 from both sides} \\ 3x<30-4 \\ 3x<26 \\ \text{ Divide both sides by 3} \\ \frac{3x}{3}<\frac{26}{3} \\ \\ \therefore x<8\frac{2}{3} \end{gathered}[/tex]- This means that all correct solutions to the inequality lie below 8.666...
- This further implies that any number greater than this is not part of the solutions of the inequality.
- 12 is greater than 8.666
Final Answer
The answer is 12
The sum of a number and -4 is greater than 15. Find the number
x > 19
Explanation:
Let the number = x
The sum of a number and -4 = x + (-4)
The sum of a number and -4 is greater than 15:
x + (-4) > 15
Multiplication of opposite signs gives negative number:
x - 4 > 15
Collect like terms:
x > 15 + 4
x > 19
Look at the figure below. 8 8 4 4 Which expression can be evaluated to find the area of this figure?
Answer
[tex]8^2-4^2[/tex]Step-by-step explanation
The figure consists of a square with sides of 8 units from which a square of sides of 4 units has been subtracted.
The area of a square is calculated as follows:
[tex]A=a^2[/tex]where a is the length of each side.
Substituting a = 8, the area of the bigger square is:
[tex]A_1=8^2[/tex]Substituting a = 4, the area of the smaller square is:
[tex]A_2=4^2[/tex]Finally, the area of the figure is:
[tex]A_1-A_2=8^2-4^2[/tex]The graph of polynomial f is shown. Select all the true statements about the polynomial.aThe degree of the polynomial is even.bThe degree of the polynomial is odd.cThe leading coefficient is positive.dThe leading coefficient is negative.eThe constant term of the polynomial is positive.fThe constant term of the polynomial is negative.
Explanation:
From the graph,
we can see that the graph is symmetric about the y axis
Hence,
We can say that the Polynomial is even
Also, Because th opwning of the function is downwards,
Hence the leading coefficient is negative
Also we can see that the y-intercept is positive
That is when x=0, y=3
Hence,
The constant term of the polynomial is positive.
Therefore,
The final answers are OPTION A,OPTION D,OPTION E
In the figure below, m2 = 49. Find mx 1.
By definition, a Right angle is an angle that measures 90 degrees.
Complementary angles are those angles that add up to 90 degrees.
For this case, you can identify that the angle 1 and the angle 2 are Complementary angles, because when you add them, you get 90 degrees (a Right angle).
Knowing the above, you can set up the following equation:
[tex]m\angle1+m\angle2=90\degree[/tex]Since you know that:
[tex]m\angle2=49\degree[/tex]You can substitute this value into the equation and the solve for the angle 1 in order to find its measure. You get that this is:
[tex]\begin{gathered} m\angle1+49\degree=90\degree \\ m\angle1=90\degree-49\degree \\ m\angle1=41\degree \end{gathered}[/tex]The answer is:
[tex]m\angle1=41\degree[/tex]write in exponential form5x5x5
5 x 5 x 5 = 5^3
[tex]\begin{gathered} \\ 5x5x5=5^{3\text{ }}\text{ = 125} \end{gathered}[/tex][tex]=16^{5\text{ }}\text{ = 16 x 16 x 16 x 16 x 16 = 1,048,576}[/tex]