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gladu [14]
4 years ago
10

Maureen has 5 rolls of pennies containing 50 coins each, 3 rolls of nickels containing 40 coins each, 6 rolls of dimes containin

g 50 coins each, and 4 rolls of quarters containing 40 coins each. How much money does she have?
Mathematics
2 answers:
harkovskaia [24]4 years ago
8 0
Answer:
she has $78.5

Explanation:
1- She has 5 rolls of pennies containing 50 coins each.
Total pennies she has = 5*50 = 250 pennies
Now, we will get the amount of pennies she has in dollars as follows:
1 penny ............> 0.01 dollar
250 pennies .....> ??
Amount is dollars = 250*0.01 = $2.5 .............> I

2- She has 3 rolls of nickels containing 40 coins each
Total nickels = 3*40 = 120 nickels
Now, we will get the amount of nickels she has in dollars as follows:
1 nickel ..............> 0.05 dollar
120 nickels .........> ??
Amount in dollars = 120*0.05 = $6 ............> II

3- She has 6 rolls of dimes containing 50 coins each
Total dimes = 6*50 = 300 dimes
Now, we will get the amount of dimes she has in dollars as follows:
1 dime ...........> 0.1 dollar
300 dimes .....> ??
Amount in dollars = $30 ...............> III

4- She has 4 rolls of quarters containing 40 coins each.
Total quarters = 4*40 = 160 quarters
Now, we will get the amount of quarters she has in dollars as follows:
1 quarter ...........> 0.25 dollar
160 quarters ........> ??
Amount in dollars = $40 ..........> IV

5- Final step:
we will add I,II,III and IV to get the total amount of money she has as follows:
Total amount of money = 2.5 + 6 + 30 + 40
Total amount of money = $78.5

Hope this helps :)



Vlad1618 [11]4 years ago
8 0
The answer is $78.50
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Compute the derivative.

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At the given point, the gradient is -7 so that

-7 = a - \dfrac b{(-1)^2} \implies a-b = -7

Eliminating b, we find

(a+b) + (a-b) = 3+(-7) \implies 2a = -4 \implies \boxed{a=-2}

Solve for b.

a+b=3 \implies b=3-a \implies \boxed{b = 5}

10. Compute the derivative.

y = \dfrac{x^3}3 - \dfrac{5x^2}2 + 6x - 1 \implies \dfrac{dy}{dx} = x^2 - 5x + 6

Solve for x when the gradient is 2.

x^2 - 5x + 6 = 2

x^2 - 5x + 4 = 0

(x - 1) (x - 4) = 0

\implies x=1 \text{ or } x=4

Evaluate y at each of these.

\boxed{x=1} \implies y = \dfrac{1^3}3 - \dfrac{5\cdot1^2}2 + 6\cdot1 - 1 = \boxed{y = \dfrac{17}6}

\boxed{x = 4} \implies y = \dfrac{4^3}3 - \dfrac{5\cdot4^2}2 + 6\cdot4 - 1 \implies \boxed{y = \dfrac{13}3}

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\dfrac{x^3}3 - 2x^2 - 8x + 5 = x + 5

\dfrac{x^3}3 - 2x^2 - 9x = 0

\dfrac x3 (x^2 - 6x - 27) = 0

\dfrac x3 (x - 9) (x + 3) = 0

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Evaluate y at each of these.

A:~~~~ \boxed{x=0} \implies y=0+5 \implies \boxed{y=5}

B:~~~~ \boxed{x=9} \implies y=9+5 \implies \boxed{y=14}

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y = \dfrac{x^3}3 - 2x^2 - 8x + 5 \implies \dfrac{dy}{dx} = x^2 - 4x - 8

Evaluate the derivative at the x-coordinates of A, B, and C.

A: ~~~~ x=0 \implies \dfrac{dy}{dx} = 0^2-4\cdot0-8 \implies \boxed{\dfrac{dy}{dx} = -8}

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Then

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