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saw5 [17]
3 years ago
13

A) Write a rule for a polynomial with left end up and right end up behavior, and has a y-

Mathematics
1 answer:
Novosadov [1.4K]3 years ago
7 0

Answer:

y = x^2 + 3

Step-by-step explanation:

The equation I put has the left end up, the right end up, and has a y-intercept at (0,3).

If this answer is correct, please make me Brainliest!

You might be interested in
(x-1)(x+2)=5(x-1) divided by x-1
anyanavicka [17]
Assuming that your question is (x-1)(x+2) = [5(x-1)]/x-1

On the right side, the x-1's will cancel out, leaving you with (x-1)(x+2) = 5

expand the left side, giving you x^2 + x -2 = 5
which goes to x^2 +x -7 = 0
the possible values for x are 2.93 and -3.93. I don't think this was your question, so I'll do the other possible question that you might have been asking.

(x-1)(x+2) = 5(x-1)

divide by x-1 on both sides, leaving you with x+2=5
x+2=5
x=5-3
x=3
5 0
3 years ago
Please help! (Answer and explanation, NO LINKS!)
blsea [12.9K]

Answer: primeter is 20

Area :

Step-by-step explanation:

Height : 4

Base : 6

Prim - 6+6+4+4=20

Area - 6x4=24

8 0
3 years ago
One weekend, a newsstand sold twice as many Sunday papers as Friday papers. The Sunday paper costs $1.50, and the Friday paper c
Fiesta28 [93]
1.50s + 0.75f = 116.25
2f = s

1.50(2f) + 0.75f = 116.25
3f + 0.75f = 116.25
3.75f = 116.25
f = 116.25/3.75
f = 31...31 friday papers

2f = s
2(31) = s
62 = s <=== 62 sunday papers
8 0
3 years ago
Can someone please help me on this
choli [55]

Answer:

if you flipped one of the figures you will see that they are both the same shape and the same distance from the x and y axes.

3 0
3 years ago
Ments
maw [93]

The sector area and the arc length are 34.92 square inches and 13.97 inches, respectively

<h3>How to find a sector area, and arc length?</h3>

For a sector that has a central angle of θ, and a radius of r;

The sector area, and the arc length are:

A = \frac{\theta}{360} * \pi r^2 --- sector area

L = \frac{\theta}{360} * 2\pi r ---- arc length

<h3>How to find the given sector area, and arc length?</h3>

Here, the given parameters are:

Central angle, θ = 160

Radius, r = 5 inches

The sector area is

A = \frac{\theta}{360} * \pi r^2

So, we have:

A = \frac{160}{360} * \frac{22}{7} * 5^2

Evaluate

A = 34.92

The arc length is:

L = \frac{\theta}{360} * 2\pi r

So, we have:

L = \frac{160}{360} * 2 * \frac{22}{7} * 5

L = 13.97

Hence, the sector area and the arc length are 34.92 square inches and 13.97 inches, respectively

Read more about sector area and arc length at:

brainly.com/question/2005046

#SPJ1

8 0
1 year ago
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