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4 years ago

8

Pamela's Purse Palace received a shipment of new purses. The total cost of the shipment was $287.20. If each purse cost $35.90,

how many purses were in the shipment?

1 answer:

ANEK [815]4 years ago

8 0

The answer is 8. The work is shown here: 287.20/35.90=8

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How many thumbs down votes did the video get

lara31 [8.8K]

So you have the ratio of 9:2.

The video has 18 thumbs up votes with the ratio being thumbs up to thumbs down.

9 in relation to 18 is times 2 so it makes since to times 2 by 2.

Therefore the ratio would be 18:4 which means there would be 4 thumbs down.

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3 years ago

Please someone explain this

vampirchik [111]

Answer: 68

Step-by-step explanation: Complementary angles are angles that add up to 90. So, you need to do 90-22=68.

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3 years ago

The admission fee at a zoo is $2 for children and $4 for adults. If 1800 people enter the zoo and $6000 is collected, how many a

ipn [44]

Answer:

The number of adults is $1,200$

Step-by-step explanation:

Let

x------> the number of children

y------> the number of adults

we know that

$x+y=1800$ -------> equation A

$2x+4y=6000$ -------> equation B

Multiply equation A by $-2$

$-2(x+y)=-2*1800$ --------> $-2x-2y=-3600$ ------> equation C

Adds equation B and equation C

$2x+4y=6000\\-2x-2y=-3600\\---------\\4y-2y=6000-3600\\2y=2400\\y=2400/2\\y=1200$

4 0

3 years ago

Read 2 more answers

Two identical rubber balls are dropped from different heights. Ball 1 is dropped from a height of 196 feet, and ball 2 is droppe

Anon25 [30]

H1 (t) = 196 - 16 t-squared. / / / H2 (t) = 271-16t-squared. / / / In each function, 't' is the number of seconds after that ball is dropped. / / / Each function is only true until the first time that H=0, that is, until the first bounce. Each function becomes very complicated after that, and we would need more information in order to write it.

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3 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago