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Nana76 [90]
2 years ago
11

Joe walks on a treadmill at a constant rate. The equation below describes the relationship between t, the time he walks in hours

, and d, the distance he walks in miles. Can someone help me with the answer...I’m not sure if it’s B or c....I know it’s not A and D.

Mathematics
1 answer:
STALIN [3.7K]2 years ago
4 0

Answer: A

Step-by-step explanation:

CORRECTED QUESTION

Joe walks on a treadmill at a constant rate. The equation d=4t describes the relationship between t, the time he walks in hours, and d, the distance he walks in miles. Which of the graph best describes the relationship?

The equation d=4t is a linear equation.

When t=1, d=4 X 1 =4

When t=2, d=4 X 2 =8

The correct graph that satisfies this pair is A (as seen in the diagram)

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The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in
Marina86 [1]

Answer:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})

And when we apply the limit we got that:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1

Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

We have the following formula in order to find the sum of cubes:

\lim_{n\to\infty} \sum_{n=1}^{\infty} i^3

We can express this formula like this:

\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

If we operate and we take out the 1/4 as a factor we got this:

\lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}

We can cancel n^2 and we got

\lim_{n\to\infty} \frac{(n+1)^2}{n^2}

We can reorder the terms like this:

\lim_{n\to\infty} (\frac{n+1}{n})^2

We can do some algebra and we got:

\lim_{n\to\infty} (1+\frac{1}{n})^2

We can solve the square and we got:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})

And when we apply the limit we got that:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1

3 0
2 years ago
Barry also looks into the cost of repaying an easy access loan for $1000. The up-front cost of the loan is $3 for every $20 borr
natta225 [31]

Answer:

Barry has to pay back $1150

Step-by-step explanation:

We can set up an equation to help solve this problem

x is the amount he has to pay back

x=1000+3(1000/20)

1000/20 lets us know how many times he borrowed 20 dollars

x=1000+3(50)

x=1000+150

x=1150

Barry has to pay back $1150

3 0
2 years ago
Product A is and 8oz bottle of cough medication that's sells for $1.36. Product B is 16oz bottle of cough medication that costs
zhenek [66]

Answer:

Product B

Step-by-step explanation:

Divide the number of ounces i the bottle by the price of the bottle. Product A has a unit price of $0.17 and Product B has a unit price of $0.20. Therefore Product B has a lower unit price :))

4 0
2 years ago
Line jk passes through points j(-4,-5) and k(-6,3 if the equation of the lines is written in slope-intercept form,y=my+b what is
lidiya [134]
Written in 2-point form, the equation of the line is
  y = (y2-y1)/(x2-x1)·(x-x1) +y1
  y = (3-(-5))/(-6-(-4))·(x-(-4)) + (-5)
  y = 8/-2·(x +4) - 5
  y = -4x -21

The value of b is -21.

6 0
2 years ago
A rectangle piece of paper has a width that is 3 inches less than it's length. It is cut in half along a diagonal to create two
yarga [219]

Answer:

A. The area of the rectangle is 88 square inches.

C. The equation x² – 3x – 88 = 0 can be used to solve for the length of the rectangle.

Step-by-step explanation:

The question is incomplete. Here is the complete question.

A rectangular piece of paper has a width that is 3 inches less than its length. It is cut in half along a diagonal to create two congruent right triangles with areas of 44 square inches. Which statements are true? Check all that apply.

A.The area of the rectangle is 88 square inches.

B.The equation x(x – 3) = 44 can be used to solve for the dimensions of the triangle.

C.The equation x2 – 3x – 88 = 0 can be used to solve for the length of the rectangle.

D.The triangle has a base of 11 inches and a height of 8 inches.

The rectangle has a width of 4 inches.

If a rectangular piece of paper has a width that is 3 inches less than it's length and it is cut in half along a diagonal to create two congruent right triangles with areas of 44 square inches, the width of the rectangle will be equal to the base of the each of the triangle, and the length equal to the height of the triangle.

Let the length of the rectangular piece of paper be L. If its width is 3 inches less than it's length, then W = L-3

The base of the triangle = L-3 and the height = L

Area of the triangle = 1/2 * base * height

Area of the triangle = 1/2 (L-3)L

If the area of the congruent triangles is 44in², then;

44 = 1/2 (L-3)L

88 = (L-3)L

88 = L²-3L

L²-3L-88=0

L²-11L+8L-88=0

L(L-11)+8(L-11)=0

(L-11)(L+8)= 0

L = 11in

This shows that the height of the triangle is 11 inches.

The base of the triangle = 11-3 = 8 inches

The area of the rectangle = Length*Breadth

= 11* 8

= 88in²

Based on the calculation above, the following statements are true.

-The area of the rectangle is 88 square inches.

-The equation x² – 3x – 88 = 0 can be used to solve for the length of the rectangle.

7 1
2 years ago
Read 2 more answers
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