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2 years ago

13

Pls help 10 ptssssss NO LINKS

1 answer:

puteri [66]2 years ago

4 0

Answer:

its five points...

Step-by-step explanation:

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What is 7/8<br> expressed as a percent?<br> Enter your answer in the box.

ASHA 777 [7]

The answer is 87.5% You can check it everywhere. :)

7 0

3 years ago

Is this a true proportion ? 1/2 = 5/10 A sometimes B Yes C none of these D no

Andrews [41]

Question

Is this a true proportion ? 1/2 = 5/10 A sometimes B Yes C none of these D no

Answer:

<h3>B Yes</h3>

Step-by-step explanation:

1/2 = 5/10

semplify 5/10

1/2 = 1/2

the proportion is true

5 0

3 years ago

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In the division problem 8 ÷ 2 = 4, which is the dividend?

Alexxandr [17]

The dividend would be 8

8 0

3 years ago

Read 2 more answers

a circus charges $2.50 for kids tickets and $6 for adult tickets last week and twice as many kids attended the circus than adult

aliina [53]

Let
x-----------> the number of kids
y----------> the number of adults

we know that
2.5x+6y=$7150---------> equation 1
x=2y-------------> equation 2
substitute 2 in 1
2.5*[2y]+6y=7150-----> 5y+6y=7150----> 11y=7150-----> y=650
x=2y----> x=2*650----> 1300

the answer is
the number of kids is 1300

5 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

3 years ago