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2 years ago

Workers have packed 1,400 glasses in 7 boxes. If they pack 3 more boxes, how many glasses will they have packed in all?

Mathematics

2 answers:

11111nata11111 [884]2 years ago

8 0

1box = 1400/7 = 200

200×3=600

1400+600=2000

sdas [7]2 years ago

4 0

Answer:

2000

Step-by-step explanation:

Given :Workers have packed 1,400 glasses in 7 boxes.

To Find :If they pack 3 more boxes, how many glasses will they have packed in all?

Solution:

Workers packed no. of glasses in 7 boxes = 1400

Workers packed no. of glasses in 1 box = $\frac{1400}{7}$

Workers packed no. of glasses in 3 boxes = $\frac{1400}{7} \times 3$

= $600$

So, initially they packed 1400 glasses

If they pack 3 more boxes so, the pack 600 glasses more

So, The total no. of glasses have packed by workers = 1400+600 = 2000

Hence they have packed 2000 glasses in all.

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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.

Ben:

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

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

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

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

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 $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

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:

Whose textbook reaches the ground first and by how many seconds?

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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