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3 years ago

Which equation represents the graphed function?

Mathematics

2 answers:

Novay_Z [31]3 years ago

4 0

Answer: C is your answer.

Step-by-step explanation:

Lena [83]3 years ago

3 0

The answer is c 3/2 x - 3 = y. :)

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You pick a number that is at least 2 with numbers 1-8

Georgia [21]

Answer:

its 4

Step-by-step explanation:

This is right because you multiply once by 2. I got 2 because you multiply by the closest number to get to 4.

7 0

3 years ago

Find all the zeros of the equation x^4-6x^2-7x-6=0

rusak2 [61]

Answer:

The zeros are

$x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

Step-by-step explanation:

We have been given the equation x^4-6x^2-7x-6=0

Use rational root theorem, we have

$a_0=6,\:\quad a_n=1$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:3,\:6,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:3,\:6}{1}$

$-\frac{2}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+2$

$=\left(x+2\right)\frac{x^4-6x^2-7x-6}{x+2}\\$

$=x^3-2x^2-2x-3$

Again factor using the rational root test, we get

$=\left(x+2\right)\left(x-3\right)\left(x^2+x+1\right)$

Using the zero product rule, we have

$x+2=0:\quad x=-2\\x-3=0:\quad x=3\\x^2+x+1=0\\\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}\\\\=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

Therefore, the zeros are

$x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

4 0

3 years ago

Read 2 more answers

Raju bought a book for Rs 35.65. He gave Rs 50 to the shopkeeper. How much

bearhunter [10]

50-35.65
=15.35 Rs is the amount the boy got from the shopkeeper

4 0

3 years ago

Rob colored 1/4 of a rectangle. what is another way to name 1/4?

Morgarella [4.7K]

One-Fourth is another way

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3 years ago

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If y = 2 and x = 5, Then solve 3x + 4 + 2y A 20 B 61 C 23 D 0

Arisa [49]

Answer:

Step-by-step explanation:

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3 years ago