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2 years ago

6 divided by 3/4 using tape diagrams

Mathematics

1 answer:

LUCKY_DIMON [66]2 years ago

3 0

Answer:

Step 1: Read the entire problem. Students will find themselves reading the story numerous times. Each time will have a different purpose in the model drawing process. Step 2: Turn the question into a sentence with a space for the answer.

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The base of a ladder is placed 5 feet away from a 10-foot-high wall, so that

dezoksy [38]

Answer:a i belive not 100% sure

Step-by-step explanation:

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2 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.

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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2 years ago

a A hole in the shape of a hemisphere is carved into each face of a cube. The holes are identical and centered at the centre of

morpeh [17]

Diameter of each hemispheral holes = the diameter of a full sphere = side length of cube = 2 units.

<h3>What is a Hemisphere?</h3>

A hemishpere is half of a sphere. Thus, two hemispheres makes one full sphere.

Given the image as shown below, since the identical hemispheral holes are centered at the center of each fce and they touch each other at only one point, it means that two hemisphere meet to make one full sphere.

Therefore, the diameter of a full sphere = side length of cube.

Thus, diameter of each hemispheral holes = the diameter of a full sphere = side length of cube = 2 units.

Learn more about hemisphere on:

brainly.com/question/9280450

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1 year ago

Hi help rn plss this is a system of equations question

otez555 [7]

Answer:

x=6, y=2

Step-by-step explanation:

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