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3 years ago

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:

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