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djverab [1.8K]
3 years ago
14

Samantha would like to purchase a new pair of shoes that cost $80.00. If she has a coupon for 25% of the price of the shoes, wha

t will be the new price of the shoes after the discount?
A. $20.00
B. $4.00
C. $60.00
D. $18.00
Mathematics
1 answer:
frutty [35]3 years ago
5 0

Answer:

C.$60.00

Step-by-step explanation:

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Julianne opens a dance studio. Her start-up costs for the building, advertising, and supplies $52,000. Each day, she spends $650
Yuki888 [10]

The equation is:  960d-650d=52000 and Julianne will begin making a profit after 168 days.

<em><u>Explanation</u></em>

Her total start-up cost is $52,000.

Each day, she spends $650 on operating costs (like utilities and wages) and she earns $960 per day from her students' lesson fees.

If the number of days to overcome the start-up cost is  d , then

the total operation cost spent in  d days = \$650d and

the total students' lesson fees earned in  d days = \$960d

<em><u>Part A:</u></em>   The equation to represent the situation will be:   960d-650d=52000  

<u><em>Part B:</em></u><em> </em><em>  </em>Julianne will begin making a profit when....

960d-650d>52000\\ \\ 310d>52000\\ \\ d>\frac{52000}{310}\\ \\ d >167.741... \approx 168

So, Julianne will begin making a profit after 168 days.

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3 years ago
Help me PLZ with the exact answer
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Y = 4/-3 - 1

The slope can either be -4/3 or 4/-3 and the y-intercept is (0,1)
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3 years ago
I will givee Brainlist if anyone answers all these questions
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What percent of 128 is 32
bija089 [108]
We can set up a proportion.

32/128 = x/100

32 * 100 = 3200
128 * x = 128x.

Now we have 128x = 3200.

All we need to do is divide 3200 by 128, so 3200 / 128 = 25, so your answer is 25%.
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2 years ago
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Makovka662 [10]

Let u=x^2-4 and v=4x-5. By the product rule,

\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=\dfrac{\mathrm d(u^5)}{\mathrm dx}v^4+u^5\dfrac{\mathrm d(v^4)}{\mathrm dx}

By the power rule, we have (u^5)'=5u^4 and (v^4)'=4v^3, but u,v are functions of x, so we also need to apply the chain rule:

\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}

\dfrac{\mathrm d(v^4)}{\mathrm dx}=4v^3\dfrac{\mathrm dv}{\mathrm dx}

and we have

\dfrac{\mathrm du}{\mathrm dx}=2x

\dfrac{\mathrm dv}{\mathrm dx}=4

So we end up with

\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3

Replace u,v to get everything in terms of x:

\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3

We can simplify this by factoring:

10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3=2(x^2-4)^4(4x-5)^3\bigg(5x(4x-5)+8(x^2-4)\bigg)

=2(x^2-4)^4(4x-5)^3(28x^2-57)

7 0
3 years ago
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