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Dima020 [189]
3 years ago
13

If you have 14 1/2 dozen boxes of envelopes, and you order 3 1/4 dozen more, how many dozen boxes will you have in all?

Mathematics
1 answer:
Elanso [62]3 years ago
5 0

The answere is 213 boxes

If you add 3 dozens to 14 dozens of boxes of envilopes , then you get 17 dozens of boxes of envilopes.Then of you sum up the 1/2 (2/4 ) with the 1/4.You get 3/4. Finally, you add 17 dozens to 3/4 of a dozen you get 17 3/4 dozens( wich is 213 boxes )

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Solve the system of equations.<br><br><br> −5x+2y=9<br> y=7x<br> ​
RUDIKE [14]

Answer:

x=1  y=7

Step-by-step explanation:

−5x+2y=9

y=7x

Substitute the second equation into the first

-5x +2(7x) = 9

Distribute

-5x +14x = 9

Combine like terms

9x=9

Divide by 9

9x/9 =9/9

x= 1

Now find y

y = 7(1)

y = 7

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Help Please 20 POINTS <br> Which expressions are equivalent to the following?
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Answer:

\large\boxed{5x(6x-x-2)}\\\\\boxed{-5(-6x^2+x+2)}\\\\\boxed{5(2x+1)(3x-2)}

Step-by-step explanation:

30x^2-5x-10=5(6x^2-x-2)=5(-1)(-6x^2+x+2)=-5(-6x^2+x+2)\\\\30x^2-5x-10=30x^2+15x-20x-10=15x(2x+1)-10(2x+1)\\\\=(2x+1)(15x-10)=(2x+1)(5)(3x-2)=5(2x-1)(3x-2)

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3 years ago
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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Answer:

Step-by-step explanation:

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