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Novosadov [1.4K]
2 years ago
6

Lucy is going to walk in a fundraising event to raise money for her school band. Her mother has agreed to donate $7.50 to the sc

hool for each mile that Lucy walks, plus an additional $25. If Lucy wants to make at least $85.00 from her mother’s donation, how far will she have to walk? Show your work.
Mathematics
1 answer:
makkiz [27]2 years ago
4 0

Answer:

8 miles

Step-by-step explanation:

let x= miles she walks

7.50x+25=85

subtract 25 from both

7.50x=60

divide both by 7.50

x=8

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Speed of student A is 180 ft/min
Speed of student B is 120 ft/min
Relative speed=(120+180)=300 ft/min
Distance between them is 2420 ft
Time taken for them to meet will be:
time=distance/speed
=(2420-20)/300
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A box contains two defective Christmas tree lights that have been inadvertently mixed with eight nondefective lights. If the lig
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possibility 2: defective-nondefective-defective

possibility 1: 8/10x 2/9 x 1/8 = 16/720=1/45
possibility 2: 2/10 x 8/9 x 1/8= 16/720=1/45

answer: 2/45
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PLSS HELPPP 20 POINTS
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Answer:

a

Step-by-step explanation:

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Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be
Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of t=0.6482

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

h(t)=-16t^2 +v_{o}t+s       Eqn(1).

Where:

h(t) : is the height range as a function of time

v_{o}   : is the initial velocity

s     : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

h(t)=-16t^2 + v_{o}t + 40        Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that v_{o}=60ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

We know that v_{o}=48ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

h(t)=0-16t^2+60t+40=0

Using the quadratic expression to find the roots of the quadratic we have:

t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec

Since time can only be positive we reject the t_{2} solution and we keep that Ben's book took t=4.3276 seconds to reach the ground.

Similarly solving for Josh we obtain

t_{1}=3.6794sec\\t_{2}=-0.6794sec

Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

  • Josh's textbook reached the ground first
  • Josh's textbook reached the ground first by a difference of t=0.6482

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