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2 years ago

10

Danielle opens a credit card with an APR of 22.79% compounded monthly. How much is charged in interest this month if her balance

is $1550?

1 answer:

salantis [7]2 years ago

8 0

1550 * 0.2279 / 12 ( monthly interest is 1/12 of APR)

= $29.43

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A construction worker has 1,032 nails. He needs to put them in containers with 12 in each container. How many containers will he

miv72 [106K]

Answer:

86

Step-by-step explanation:

1,032 ÷ 12 = 86

86 containers would be needed.

4 0

1 year ago

3) Let h(x)= |2x|-4. Find h(x) when x = -7.<br> a) -10<br> b) 10<br> c) -13<br> d) 13

svlad2 [7]

Answer:

$h(x) = 10$

Step-by-step explanation:

Given

$h(x) = |2x| - 4$

$x = -7$

Find h(x)

Substitute -7 for x

$h(x) = |2*-7| - 4$

$h(x) = |-14| - 4$

Absolute values return positive. So:

$h(x) = 14 - 4$

$h(x) = 10$

8 0

1 year ago

What is 1/4 + 1/4 + 1/4 =

Zepler [3.9K]

3/4 is the answer

your welcome

3 0

2 years ago

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An 8 by 19 ft wall has three 4 by 5ft windows. The area of three windows is what percent of the area of the entire wall

sergij07 [2.7K]

Area of the wall = 8 x 19 = 152 ft square
Area of the window = 4 x 5 = 20 ft square
Area of 3 windows = 20 x 3 = 60 ft square
The percent area of 3 windows from area of the wall = 60/152 or approximately .395 or 39.5%

6 0

2 years ago

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

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

I gotta go so I'll stop here, sorry.

7 0

2 years ago

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