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Alinara [238K]
4 years ago
5

Consider the linear function that is represented by the equation y = 2 x + 2 and the linear function that is represented by the

graph below.
Which statement is correct regarding their slopes and y-intercepts?

A. The function that is represented by the equation has a steeper slope and a greater y-intercept.

B. The function that is represented by the equation has a steeper slope, and the function that is represented by the graph has a greater y-intercept.

C. The function that is represented by the graph has a steeper slope, and the function that is represented by the equation has a greater y-intercept.

D. The function that is represented by the graph has a steeper slope and a greater y-intercept.

Mathematics
1 answer:
frutty [35]4 years ago
3 0

Answer:

C) the function represented by the graph has a steeper slope and the function represented by the equation has a larger y-intercept.

Step-by-step explanation:

The equation is written in slope-intercept form, y=mx+b.  m is the slope and b is the y-intercept.  In this function, the slope is 2 and the y-intercept is 2.

Looking at the graph, the y-intercept is 1.  Going from the y-intercept, if we go up 2, it does not go over quite 1 unit.  Therefore the slope is steeper on the graph.

Hope I helped answer the question:)

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Mark records his science scores in each monthly assessment over a period of 5 months. In the first assessment he scores 76%. In
katovenus [111]

d. both a relation and a function:

Given:

Mark records his science scores in each monthly assessment over a period of 5 months. In the first assessment he scores 76%. In the second assessment he scores 73%. After that, his scores keep increasing by 2% in every assessment.

x represents the number of assessments since he starts recording and y represents the scores in each assessment.

In order for a relation to be a function the association has to be unambiguous that means that for a given input only one output can exist.If an input can have two or more outputs then you cannot determine which is the correct output for that input.

In the given situation:

x is the input that is number of assessments since mark starts recording the scores so there is only one assessment no repeating.so there is only one output.

Hence the relation is a function.

Learn more about the function here:

brainly.com/question/5975436

#SPJ1

4 0
1 year ago
7.
anzhelika [568]
The answer is D as said above
8 0
3 years ago
during a sale , a shop allows discount of the marked price of clothing . What will a costumer pay for a dress with a marked pric
ollegr [7]

Answer:

35 d

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Given: KLMN is a parallelogram, KA − angle bisector of ∠K LA − angle bisector of ∠L Prove: m∠KAL = 90°
sineoko [7]

Answer: angle KAL = 90°

Step-by-step explanation: The decision on a photo

6 0
4 years ago
use the general slicing method to find the volume of The solid whose base is the triangle with vertices (0 comma 0 )​, (15 comma
lyudmila [28]

Answer:

volume V of the solid

\boxed{V=\displaystyle\frac{125\pi}{12}}

Step-by-step explanation:

The situation is depicted in the picture attached

(see picture)

First, we divide the segment [0, 5] on the X-axis into n equal parts of length 5/n each

[0, 5/n], [5/n, 2(5/n)], [2(5/n), 3(5/n)],..., [(n-1)(5/n), 5]

Now, we slice our solid into n slices.  

Each slice is a quarter of cylinder 5/n thick and has a radius of  

-k(5/n) + 5  for each k = 1,2,..., n (see picture)

So the volume of each slice is  

\displaystyle\frac{\pi(-k(5/n) + 5 )^2*(5/n)}{4}

for k=1,2,..., n

We then add up the volumes of all these slices

\displaystyle\frac{\pi(-(5/n) + 5 )^2*(5/n)}{4}+\displaystyle\frac{\pi(-2(5/n) + 5 )^2*(5/n)}{4}+...+\displaystyle\frac{\pi(-n(5/n) + 5 )^2*(5/n)}{4}

Notice that the last term of the sum vanishes. After making up the expression a little, we get

\displaystyle\frac{5\pi}{4n}\left[(-(5/n)+5)^2+(-2(5/n)+5)^2+...+(-(n-1)(5/n)+5)^2\right]=\\\\\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2

But

\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2=\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}((5/n)^2k^2-(50/n)k+25)=\\\\\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)

we also know that

\displaystyle\sum_{k=1}^{n-1}k^2=\displaystyle\frac{n(n-1)(2n-1)}{6}

and

\displaystyle\sum_{k=1}^{n-1}k=\displaystyle\frac{n(n-1)}{2}

so we have, after replacing and simplifying, the sum of the slices equals

\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)=\\\\=\displaystyle\frac{5\pi}{4n}\left(\displaystyle\frac{25}{n^2}.\displaystyle\frac{n(n-1)(2n-1)}{6}-\displaystyle\frac{50}{n}.\displaystyle\frac{n(n-1)}{2}+25(n-1)\right)=\\\\=\displaystyle\frac{125\pi}{24}.\displaystyle\frac{n(n-1)(2n-1)}{n^3}

Now we take the limit when n tends to infinite (the slices get thinner and thinner)

\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}\displaystyle\frac{n(n-1)(2n-1)}{n^3}=\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}(2-3/n+1/n^2)=\\\\=\displaystyle\frac{125\pi}{24}.2=\displaystyle\frac{125\pi}{12}

and the volume V of our solid is

\boxed{V=\displaystyle\frac{125\pi}{12}}

3 0
3 years ago
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