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3 years ago

Distributive prperty m(13-2r)

Mathematics

1 answer:

exis [7]3 years ago

5 0

Answer:

13m-26mr boom that's it

Step-by-step explanation:

i think

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How tall is the green character?

vodomira [7]

My guess......... um 5'4

5 0

3 years ago

Which term best completes the state all _______ are rectangles A. Parallelograms B.squares C.rhombi D. None of theses

liq [111]

Answer:

The answer is B. squares.

Step-by-step explanation:

All squares are considered rectangles. Not all parallelograms are rectangles, and rhombi are not rectangles.

I hope this helped! :-)

8 0

3 years ago

Susanna has played the piano for s years. Patrick has played the piano for 4 more than twice the number of years that Susanna ha

scoray [572]

S=Years Susanna has played the piano
p=Years Patrick has played the piano

The expression for this question would be s(2)+4=p

Hope this helps! :)

6 0

2 years ago

Read 2 more answers

Determine the domain of the function (fog)(x) where f(x)=3x-1/x-4 and g(x)=x+1/x

Katarina [22]

Answer: (-∞,-1) ∪ (0,+∞)

Step-by-step explanation: The representation fog(x) is a representation of composite function, meaning one depends on the other.

In this case, fog(x) means:

fog(x) = f(g(x))

fog(x) = $3(x+\frac{1}{x} )-\frac{1}{x+\frac{1}{x} } -4$

$fog(x)=3x+\frac{3}{x} -\frac{1}{\frac{x^{2}+x}{x} } -4$

$fog(x)=3x+\frac{3}{x} -\frac{x}{x^{2}+x} -4$

$fog(x)=\frac{3x^{2}(x^{2}+x)+3(x^{2}+x)-x-4x(x^{2}+x)}{x(x^{2}+x)}$

$fog(x)=\frac{3x^{4}+3x^{3}+3x^{2}+3x-x-4x^{3}+4x^{2}}{x(x^{2}+x)}$

$fog(x)=\frac{3x^{4}-x^{3}-x^{2}+2x}{x(x^{2}+x)}$

This is the function fog(x).

The domain of a function is all the values the independent variable can assume.

For fog(x), denominator can be zero, so:

$x(x^{2}+x) \neq 0$

If x = 0, the function doesn't exist.

$x^{2}+x \neq0$

$x(x+1) \neq0$

$x+1\neq0$

$x\neq-1$

Therefore, the domain of this function is: -∞ < -1 or x > 0

5 0

3 years ago

Rewrite the expression in the form 2". E

Vlad [161]

Answer:

z^ {8/5}

Step-by-step explanation:

$(z^{\frac{-4}{3})^{\frac{-6}{5} } \\We\ know\ that\ a^{(b)^c} = a^{bc}\\\\Hence, \\\\{z^{\frac{-4}{3})^{\frac{-6}{5}}\\}={z^{\frac{-4}{3}*\frac{-6}{5}\\}\\=z^{\frac{24}{15}}\\=z^\frac{8}{5}$

6 0

2 years ago