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SVEN [57.7K]
3 years ago
9

It takes about 14 person-hours of labor to break up a driveway and clear away the rubble. It takes about 8 person-hours of labor

to pour and smooth concrete for the new driveway.
Mathematics
1 answer:
MAVERICK [17]3 years ago
8 0

Answer: 22 person-hours

Step-by-step explanation:

It takes 14 hours to break up the driveway and then to clear away the rubble.

Then it takes 8 hours to pour the concrete and smooth it.

Total time it will take to break up a driveway, clear away the rubble, pour the concrete, and smooth the concrete will therefore be;

= 14 + 8

= 22 person-hours

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a model used to compare and contrast

Step-by-step explanation:

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Mrs. Wright bought some notebook paper for her class. She decided to keep 3/4 of the paper for her own use. How much paper did s
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So the total amount of paper is 4/4 right?
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before Christmas a story steals fake Christmas trees for $289 after Christmas they sell the same trees for $150 what is the appr
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6 0
3 years ago
5x - 4y = 19 <br> x + 2y = 8
klemol [59]

Answer:

(5x-4y)=19

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95-76

19

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7 0
3 years ago
Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the
Rama09 [41]

<em><u>Question:</u></em>

Britney throws an object straight up into the air with an initial velocity of 27 ft/s from a platform that is 10 ft above the ground. Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the object to hit the ground?

1s   2s   3s   4s

<em><u>Answer:</u></em>

It takes 2 seconds for object to hit the ground

<em><u>Solution:</u></em>

<em><u>The given equation is:</u></em>

h(t) = -16t^2 + v_0t+h_0

Initial velocity = 27 feet/sec

h_0 = 10\ feet

Therefore,

h(t) = -16t^2 +27t+10

At the point the object hits the ground, h(t) = 0

-16t^2 +27t+10 = 0\\\\16t^2-27t-10=0

Solve by quadratic formula,

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=16,\:b=-27,\:c=-10:\quad \\\\t=\frac{-\left(-27\right)\pm \sqrt{\left(-27\right)^2-4\cdot \:16\left(-10\right)}}{2\cdot \:16}\\\\t = \frac{27 \pm \sqrt{1369}}{32}\\\\t = \frac{27 \pm 37}{32}\\\\We\ have\ two\ solutions\\\\t = \frac{27+37}{32}\\\\t = \frac{64}{32}\\\\t = 2\\\\And\\\\t = \frac{27-37}{32}\\\\t = -0.3125

Ignore, negative value

Thus, it takes 2 seconds for object to hit the ground

7 0
3 years ago
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