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Rama09 [41]
3 years ago
15

What kind of polynomial has a degree of 2? Cubic Binomial Quadratic Linear

Mathematics
1 answer:
Kaylis [27]3 years ago
6 0

Answer:

Quadratic Function

Step-by-step explanation:

A quadratic Function has a degree of 2


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Help pls asap!! No links.
algol [13]

Answer:

Step-by-step explanation:

1. Do what is inside the parantheses

4^{2} +3+2\\16+3+2 = 21

7^{2} +3(21)

2. Simplify both terms

7^{2} =49\\3(21) = 63\\49+63

3. Solve

49+63 = 112

7 0
2 years ago
Read 2 more answers
Todd has a collection of 8 CDs and Sally has a collection of 18 CDs. Todd is adding 9 CDs a month to his collection while Sally
Anuta_ua [19.1K]

todd = 8 + 9

sally = 18 + 7

Just keep adding until they have the same amount if they ever do.

todd 17 26 35 44 53

sally 25 32 39 46 53


so it took them  5 mouths and they both have 53  CD's

5 0
3 years ago
PLEASE HELP 10 POINTS
abruzzese [7]
A) 70 pools in total
B) 5:2
5 0
3 years ago
Find the measure of the arc or angle indicated
mr_godi [17]

Answer:

b

Step-by-step explanation:

6 0
3 years ago
Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,
torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

y=6x^2+14x

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

\frac{dy}{dx}=12x+14

We know that length of curve

s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx

We have a=-2 and b=1

Using the formula

Length of curve=s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

dt=12dx

dx=\frac{1}{12}dt

Length of curve=s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt

We know that

\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C

By using the formula

Length of curve=s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}

Length of curve=s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}

Length of curve=s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})

Length of curve=s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})

Length of curve=s=32.66

5 0
3 years ago
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