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jolli1 [7]
2 years ago
11

Tyler further earns $12 per hour the store offers him a raise a 5% increase per hour after the race how much would Tyler‘s broth

er make per hour
Mathematics
1 answer:
Dmitry [639]2 years ago
7 0

Answer:

Step-by-step explanation:

First calculate the amount his hourly rate will increase by. So multiply 13 by 5% which is $12*0.05=0.60. Then add this amount to his current hourly rate. So he now makes $12.60

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zvonat [6]
They would cost $8.75. If you take out 30% from the original price, that is how much they would be.
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3 years ago
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Find the ordered pair that makes both the equations below true
NNADVOKAT [17]

Answer:

(3, 1)

Step-by-step explanation:

(a) Algebraic solution

(1)  y = -⅔x + 3

(2) y =   2x  - 5

Set Equation (1) equal to Equation (2)

-⅔x + 3 = 2x - 5

Multiply each side by 3

-2x + 9 = 6x - 15

Add 15 to each side

-2x + 24 = 6x

Add 2x to each side

24 = 8x

Divide each side by 3

(3) x = 3

Substitute (3) into (2)

y = 2×3 - 5 = 6 - 5 = 1

The ordered pair that makes both equations true is (3, 1).

(b) Graphical solution

In the diagram below, the red line is the graph of Equation (1). The blue line is the graph of Equation (2). The point of intersection is at (3, 1).

7 0
3 years ago
Which is the value of 3b2−b3b2-b when b = 5?
kow [346]
What's the equation exactly?

8 0
3 years ago
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If 33 laptops cost 30,000, what proportion could be used to determine the cost of 11 laptops?
andre [41]

Answer:

30,00 divided by 33= 909.09 times 11=9,999.99  hope it helps you

Step-by-step explanation:

6 0
2 years ago
The function in Exercise represents the rate of flow of money in dollars per year. Assume a 10-year period at 8% compounded cont
Kryger [21]

Answer:

a) P= $56,972.5

b) A=$1,26,792.3

Step-by-step explanation:

Given Data:

Interest rate=r= 0.08  per year

No. of years=t=10

Rate of continuous money flow is given by the function

f(t)=2000

a) to find the present value of money

P=\int\limits^n_0 {f(t)e^{-rt} } \, dt

Put f(t)=2000 and n=10 years and r=0.08

P=\int\limits^n_0 {2000e^{-0.08t} } \, dt

Now integrate

P= {2000(\frac{e^{-0.08t}}{-0.08} )

P= -\frac{2000}{0.08} (e^{-0.08*10}-e^{-0.08*0})

P= -\frac{2000}{0.08} (e^{-0.8}-e^{0})

P= -\frac{2000}{0.08} (0.4493-2.7282)

P= -\frac{2000}{0.08} (-2.2789)

P= -25000(-2.2789)

P= $56972.5

(b) to find the accumulated amount of money at t=10

A=P(e^{rt} )

Where P is the present worth already calculated in part a

A=56972.5(e^{0.08*10} )

A=56972.5(e^{0.8} )

A=56972.5(2.2255 )

A=$1,26,792.3

5 0
3 years ago
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