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anyanavicka [17]
2 years ago
7

Question 3 of 10 What is the solution to this equation? 4(x-8) + 10 = -10 O A. x= 3 OB. x = 5 O c. x=2 O D. x= 4​

Mathematics
1 answer:
krek1111 [17]2 years ago
5 0

Answer:

O A. x = 3

Step-by-step explanation:

4(x - 8) + 10 = -10

4x - 32 + 10 = -10

4x = 12

x = 3

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What is the length of the third side of the window frame below?
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Answer:

15 inches

Step-by-step explanation:

The longest side of the right triangular window frame is 39 inches

The height is 36 inches

Let the base of the window frame be x inches

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x² = 39² - 36² = 225

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The third side of the window frame is therefore equal to 15 inches.

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A jar contains only​ pennies, nickels,​ dimes, and quarters. There are 16 ​pennies, 18 ​dimes, and 28 quarters. The rest of the
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2 years ago
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Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

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3 years ago
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CALC- limits<br> please show your method
gladu [14]
A. Factor the numerator as a difference of squares:

\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6

c. As x\to\infty, the contribution of the terms of degree less than 2 becomes negligible, which means we can write

\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4

e. Let's first rewrite the root terms with rational exponents:

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Answer:

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