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3 years ago

15

Marcia will make a frame for her picture the length of the picture is 15 inches the width is one third of the length how much wo

od does marcia need for

1 answer:

vovangra [49]3 years ago

4 0

The answer to that would be 5 !!!!

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Omar was born on December 1, 1998, Omar's mom was born on January 18, 1965.

worty [1.4K]

Answer:

33 yrs old

Step-by-step explanation:count up to the year he was born

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3 years ago

4 (X+7) = 38<br> WHAT IS X AND EXPLAIN HOW YOU KNOW

xxMikexx [17]

Answer:

X=2.5

Step-by-step explanation:

38÷4 = 9.5

9.5 - 7=2.5

Divide each side by four to get rid of that numbr from the elft and subtract seven from both sides to get rid of the seven so your left with x=2.5

7 0

2 years ago

Read 2 more answers

Alysia learns that when she opens a new bag of candy, she has a 2/5 chance of pulling out a piece of ridiculous raspberry and a

Arlecino [84]

2/5=6/15
1/3=5/15
11/15

yes. that would be the last one possible.

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3 years ago

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18) In an isosceles triangle, the altitude and median from the vertex angle are the same line as the bisector of the vertex angl

Minchanka [31]

The answer is: ALWAYS TRUE .
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8 0

2 years ago

Calculate the total area of the shaded region.

LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

$\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}$

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

$\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill$

$\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}$

7 0

2 years ago