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2 years ago

I NEED HELP PLS I FORGOT HOW TO DO THIS HELP

Mathematics

2 answers:

KengaRu [80]2 years ago

7 0

Answer:

Step-by-step explanation:

how many shapes of size A are in B ?

I counted 9.5

then just multiply the count * the size of A

got it?

nikklg [1K]2 years ago

6 0

<h2><u>Answer</u>: 28.5cm²</h2><h2>_____________________</h2>

Step-by-step explanation:

area of shape A = 3cm²

number of A shape in shape B = 9 and half

1 shape A = 3cm²

9 shape A = 3×9 = 27cm²

1 half shape A = 3 ÷ 2 = 1.5cm²

area of shape B = 27cm² + 1.5cm² = 28.5cm²

<h2>_____________________</h2><h2>Follow me</h2>

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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

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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).

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$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

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

Thus Eqn. (2) becomes:

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

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
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Answer:

Check Explanation

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The days of the week include

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Monday

Tuesday

Wednesday

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All the letters that cannot be taught using this method because no day of the week contains them include

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c) In Tuesday, there are 7 distinct letters.

So, the sample space consist of 7 letters.

n(sample space) = 7

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Probability of picking a Y = (1/7)

The reasoning is simple, picking a Y on a Tuesday is one out of 7 possible outcomes, so, the probability of picking a Y on a Tuesday to teach the kid about = (1/7) = 0.143

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