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3 years ago

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Sarah took the advertising department from her company on a round trip to meet with a potential client. Including Sarah a total

of 11 people took the trip. She was
able to purchase coach tickets for $330 and first class tickets for $1110. She used her total budget for airfare for the trip, which was $8310. How many first class
tickets did she buy? How many coach tickets did she buy?
number of first class tickets bought

2 answers:

oksano4ka [1.4K]3 years ago

6 0

Answer:

6 first class 5 coach

Step-by-step explanation:

msg me if you need me to explain it

lesya [120]3 years ago

3 0

Answer: 8 First class tickets

Step-by-step explanation:

You can set up a system of equations for this problem. Let x = number of coach tickets and y = number of first class tickets. Then:

330x + 1220y = 12730 (cost of coach tickets plus cost of first class tickets is total budget)

x + y = 17 (number of coach tickets plus number of first class tickets is total number of people)

Solve the second equation for y to get y = 17 - x, then plug that into the first equation and solve for x:

330x + 1220(17 - x) = 12730

330x + 20740 - 1220x = 12730

-890x + 20740 = 12730

-890x = -8010

x = 9

Sarah bought x = 9 coach tickets. Plug that into the second equation and solve for y:

9 + y = 17

y = 8

Sarah bought y = 8 first class tickets.

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Taya2010 [7]

Answer

Josh's textbook reached the ground first

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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:

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$v_{o}$ : 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:

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<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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