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2 years ago

Given f(x) = 3x + 4 and g(x) = 8x + 1 then what is ( f + g )(0))? *****HELP*^^^^^^

1 answer:

3 0

• f(x) = 3x + 4

• g(x) = 8x + 1

Sum both functions above:

(f + g)(x) = f(x) + g(x)

(f + g)(x) = (3x + 4) + (8x + 1)

(f + g)(x) = 3x + 8x + 4 + 1

(f + g)(x) = 11x + 5

Therefore,

(f + g)(0) = 11 · 0 + 5

(f + g)(0) = 0 + 5

(f + g)(0) = 5 <——— this is the answer.

I hope this helps! =)

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On a coordinate plane, 2 straight lines are shown. The first solid line has a positive slope and goes through (0, negative 2) an

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Answer:

$y\geq x-2$

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Step-by-step explanation:

step 1

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Remember that

Everything to the left of the solid line is shaded

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step 2

Find the equation of the second inequality

Find the equation of the second dashed line

we have the ordered pairs

(0,2) and (4,0)

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The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

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The equation in slope intercept form is equal to

$y=mx+b$

we have

$m=-\frac{1}{2}$

$b=2$ ---> the y-intercept is given

substitute

$y=-\frac{1}{2}x+2$

Remember that

Everything below and to the left of the line is shaded

The inequality is

$y$

Rewrite

Multiply by 2 both sides

$2y$

$x+2y$

therefore

The system of inequalities is

$y\geq x-2$

$x+2y$

see the attached figure to better understand the problem

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3 years ago

Read 2 more answers

The length of an edge of a cube is 2 inches. How many cubic inches are there in the volume of the cube ?

gizmo_the_mogwai [7]

A cube has equal edges, so 2*2*2 or 2^3 (2 cubed) v=8^3

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3 years ago

How to prove this???

swat32

$\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 
$

go to right side now

$4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)$

use $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$ and $x^3 + y^3 = x^2 - xy + y^2$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)$

so $\sin^2 A + \cos^2 A = 1$

$4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side$

4 0

2 years ago