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2 years ago

WHY IS NOBODY HELP ME I GOT 2 OTHER QUESTIONS POSTED AND NOBODY ANSWERED ME

Mathematics

2 answers:

lina2011 [118]2 years ago

8 0

Answer:

B. F(0) = 4

Step-by-step explanation:

Im unsure about the C tho.

andriy [413]2 years ago

3 0

Not everybody knows the answers and sees your questions

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A prism has a volume of 56. If the dimensions of the prism are tripled, what is

defon

Answer:

A. 1512

Step-by-step explanation:

Ratio of Volumes = (Ratio of sides)³

V/56 = 3³

V/56 = 27

V = 56 × 27

V = 1512

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3 years ago

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Helppppppppppppppppppp

ryzh [129]

Answer:

Step-by-step explanation:

You just substitute the values in for A and B and then you simplify

The 10 A would be 10x17 which is 170 and then plus B, which we know is 12, so it would equal 182

hope this helped

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3 years ago

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Angle XYZ is bisected by line YW the measure of angle XYW is 52 what is the measurement of angle WYZ what is the measurement of

Aliun [14]

Answer:

Step-by-step explanation:

2(2x+7) = 6x-8

4x + 14 = 6x - 8

2x = 22

x = 11

m<XYZ = 58

m<XYW = 29

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

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Finding the expression

vekshin1

If you combine like terrms its easy, you just have to -5+5+6y, 6y=16 y=2.66666667

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3 years ago