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2 years ago

7

I WILL GIVE BRAINLIEST TO PERSON WHO ANSWERS

2 answers:

Lorico [155]2 years ago

6 0

Answer:

The Correct Answer is -50

Step-by-step explanation:

Leviafan [203]2 years ago

4 0

Answer:

-50º

Step-by-step explanation:

0 - 50 = -50

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PLZ HELP ASAP!!<br><br>I got the quadrant part I just need help with 0ther parts

Gnesinka [82]

Answer:

Step-by-step explanation:

The X intercept is 1 and the y-intercept is 1 aswell.

7 0

3 years ago

Read 2 more answers

Translate algebraic expression. <br><br>1) One-fourth of the sum of 6 and s?

zhenek [66]

1/4 (6 + s)

One-fourth (1/4) of the sum (+) of: 6 and s (6 + s)

3 0

3 years ago

Some cows are brown. Fido is not a cow. Fido must be brown.

Lina20 [59]

Answer:

B.) Invalid Argument

Step-by-step explanation:

Just because Fido is not a cow does not mean that he will automatically be brown. It is already stated that some cows(something Fido is not) are colored brown.

8 0

3 years ago

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What is the minimum number of points that could be moved to make the parallelogram a<br> rectangle?

scoray [572]

Answer:

2

Step-by-step explanation:

If we have the length and the breadth of a rectangle then we can draw a unique rectangle because we know already that all the vertex angles are equal to 90 degrees. Hence, only two measurements are required.

4 0

3 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

4 years ago