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3 years ago

8

a store marks up sporting goods 27%. write an expression that equals the retail price of an item with an original cost of p doll

ars?

1 answer:

tresset_1 [31]3 years ago

5 0

P times .27 plus p equals rp (retail Price)

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A taxi driver had 27 fares to and from the airport last Monday. The price for a ride to the airport is $14, and the price for a

OLga [1]

Answer:

(15, 12)

Step-by-step explanation:

Let's generate two systems of equations that fit this scenario.

Number of trips to the airport = x

Number of trips from the airport = y

Total number of trips to and from the airport = 27

Thus:

$x + y = 27$ => equation 1.

Total price for trips to the Airport = 14*x = 14x

Total price of trips from the airport = 7*y = 7y

Total collected for the day = $294

Thus:

$14x + 7y = 294$ => equation 2.

Multiply equation 1 by 7, and multiply equation 2 by 1 to make both equations equivalent.

7 × $x + y = 27$

1 × $14x + 7y = 294$

Thus:

$7x + 7y = 189$ => equation 3

$14x + 7y = 294$ => equation 4

Subtract equation 4 from equation 3

-7x = -105

Divide both sides by -7

x = 15

Substitute x = 15 in equation 1

$x + y = 27$

$15 + y = 27$

Subtract both sides by 15

$y = 27 - 15$

$y = 12$

The ordered pair would be (15, 12)

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3 years ago

Pretend that you are a strawberry farmer and the market price of strawberries went up $1 per pound to $4 per pound. Explain what

Karolina [17]

Increase the price of the strawberries to keep the same profit

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4 years ago

This figure is made up of two rectangular prisms.

Crazy boy [7]

3•2•7=42•2=84
12•10•7=840
84+840=924in^3

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3 years ago

Read 2 more answers

17 < 2x + 7 <29*<br> Your Inswer

borishaifa [10]

Answer:

5<x<11

Step-by-step explanation:

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3 years ago

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A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

3 years ago