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3 years ago

What are the roots of the function? f(x) =x^2 + 2x – 80

Mathematics

1 answer:

Andrej [43]3 years ago

7 0

Answer:

x^2+2x-8= x^2+10x-8x-80 =x(x+10)-8(x+10)

=( x-8) (x+10)

therefore x= 8 or x=-10

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PLEASE HELP ME! ASAP!! I’m begging you!!

ANTONII [103]

Answer:

pg ==15

Step-by-step explanation:

pg=15

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3 years ago

Once upon a time, on an island in the middle of an ocean, several tribes lived and fought with each other. A time came when all

n200080 [17]

Answer:

4 tribes. I don't even know how to do this question at all since it's way too difficult for me but this is an attempt.

Step-by-step explanation:

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3 years ago

Lowering powers write in terms of first power of cosine. Cos^6

yaroslaw [1]

The main identity you need is the double angle one for cosine:

$\cos^2x=\dfrac{1+\cos2x}2$

We get

$\cos^6x=(\cos^2x)^3=\left(\dfrac{1+\cos2x}2\right)^3=\dfrac{(1+\cos2x)^3}8$

Expand the numerator to apply the identity again:

$\cos^6x=\dfrac{1+3\cos2x+3\cos^22x+\cos^32x}8$

$\cos^6x=\dfrac{1+3\cos2x+3\left(\frac{1+\cos2(2x)}2\right)+\cos2x\left(\frac{1+\cos2(2x)}2\right)}8$

$\cos^6x=\dfrac{1+3\cos2x+\frac32+\frac32\cos4x+\frac12\cos2x(1+\cos4x)}8$

$\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{16}\cos2x\cos4x$

Finally, make use of the product identity for cosine:

$\cos2x\cos4x=\dfrac{\cos6x+\cos2x}2$

so that ultimately,

$\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos2x+\dfrac1{32}\cos6x$

$\cos^6x=\dfrac5{16}+\dfrac{15}{32}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos6x$

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4 years ago

X=y which y=7xX what is the answer

77julia77 [94]

I think it is x=7/2.

8 0

4 years ago

2/3 written as a fraction with a denominator of 9

Usimov [2.4K]

6/9 just multiply both parts of the fraction by 3

5 0

3 years ago

Read 2 more answers