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Trava [24]
3 years ago
6

The sum of two numbers is 68 if four times the smaller number is subtracted from the larger​ number, the result is 3 find the tw

o numbers.
Mathematics
1 answer:
allochka39001 [22]3 years ago
5 0
If we say:
x = the smaller numbers we want to find
y = the larger number we want to find
Then, we can formulate two equations using the information given:
x + y = 68 [1]
y - 4x = 3 [2]
Now, we can solve simultaneously:
Rearrange [1] & [2] in terms of y:
y = 68 - x [1]
y = 4x + 3 [2]
Now equate them and solve for x:
68 - x = 4x + 3
5x + 3 = 68
5x = 65
x = 13
Sub x-value into either [1] or [2]:
y = 68 - (13)
y = 55
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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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3 years ago
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FIND (-8/3)+(-1/4)+(-11/6)+3/8-3
emmainna [20.7K]

Answer:

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Step-by-step explanation:

The least common denominator of 3, 4, 6, and 8 is 24; converting all of our fractions to that denominator and solving, we have

-\frac{8}{3}+\left(-\frac{1}{4}\right)+\left(-\frac{11}{6}\right)+\frac{3}{8}-3\\=-\frac{64}{24}+\left(-\frac{6}{24}\right)+\left(-\frac{44}{24}\right)+\frac{9}{24}-3\\\\=\frac{-64+(-6)+(-44)+9}{24} -3\\\\=\frac{-70+(-44)+9}{24} -3\\\\=\frac{-114+9}{24} -3\\\\=\frac{-105}{24}-3 \\\\=\frac{-35}{8}-3\\ \\=-4\frac{3}{8}-3\\\\=-7\frac{3}{8}

So our solution is -7 3/8

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