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3 years ago

12

(3, -2), y=x + 4 i need help to find the parallel equation. please explain how you got the answer so i know how to solve.

1 answer:

GarryVolchara [31]3 years ago

3 0

I hope this helps you

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Mary wants to find this product. 58 x 20 How can Mary find the product? A. 0 +16 + 100 B. 0 + 160 + 1000 . C. 58 + 16 + 100 D. 5

Fittoniya [83]

Answer:

B

Step-by-step explanation:

58×20=1160

0+160+1000=1160

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3 years ago

Write (2/3)^2 in expanded form,

PtichkaEL [24]

Answer:

4/9, or 0.4∞

but 0.4∞ has a repeating decimal, so the expanded form is ∞ also

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3 years ago

Two sides of a triangle have the same length. The third side measures 4 m less than twice the common length. The perimeter of th

velikii [3]

Answer:

the length of the two equal sides: 8m

the length of the third side: 8m-4m=4m

8m×2+4m=20m

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3 years ago

Is my answer right? if not can you please tell me which won is right..

cupoosta [38]

<h2><u>A = 4</u> is the correct answer!</h2><h3></h3><h3>3 x ? = 12</h3><h3>12 ÷ 3 = 4</h3><h3>so</h3><h3>1 x 4 = 4</h3><h3 /><h3>You're wrong. It is not six.</h3><h3>By the way, it's "one" not "won".</h3><h3>It was probably a mistake.</h3><h3>:)</h3><h3 /><h3><em>Please let me know if I am wrong.</em></h3>

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2 years ago

A. Use composition to prove whether or not the functions are inverses of each other. B. Express the domain of the compositions u

Kryger [21]

Given: $f(x) = \frac{1}{x-2}$

$g(x) = \frac{2x+1}{x}$

A.)Consider

$f(g(x))= f(\frac{2x+1}{x} )$

$f(\frac{2x+1}{x} )=\frac{1}{(\frac{2x+1}{x})-2}$

$f(\frac{2x+1}{x} )=\frac{1}{\frac{2x+1-2x}{x}}$

$f(\frac{2x+1}{x} )=\frac{x}{1}$

$f(\frac{2x+1}{x} )=1$

Also,

$g(f(x))= g(\frac{1}{x-2} )$

$g(\frac{1}{x-2} )= \frac{2(\frac{1}{x-2}) +1 }{\frac{1}{x-2}}$

$g(\frac{1}{x-2} )= \frac{\frac{2+x-2}{x-2} }{\frac{1}{x-2}}$

$g(\frac{1}{x-2} )= \frac{x }{1}$

$g(\frac{1}{x-2} )= x$

Since, $f(g(x))=g(f(x))=x$

Therefore, both functions are inverses of each other.

B.

For the Composition function $f(g(x)) = f(\frac{2x+1}{x} )=x$

Since, the function $f(g(x))$ is not defined for $x=0$ .

Therefore, the domain is $(-\infty,0)\cup(0,\infty)$

For the Composition function $g(f(x)) =g(\frac{1}{x-2} )=x$

Since, the function $g(f(x))$ is not defined for $x=2$ .

Therefore, the domain is $(-\infty,2)\cup(2,\infty)$

8 0

3 years ago