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otez555 [7]
3 years ago
9

Simplify the expression. Write the answer using scientific notation. (4 x 10^13)^-2 -8 x 10^-26 6.25 x 10^-27 0.0625 x 10^-26 6.

25 x 10^-28
Mathematics
2 answers:
zimovet [89]3 years ago
6 0

Answer:

Option 4th is correct that is:6.25\cdot 10^{-28}

Step-by-step explanation:

We have been given an expression:

(4\cdot 10^{13})^{-2}

4^{-2}\cdot 10^{13\cdot -2}

Using x^{-a}=\frac{1}{x^a}

Here, x= and a= -2

\frac{1}{4^2}\cdot 10^{-26}

On simplification:

0.0625\cdot10^{-26}

When we right shift the decimal to 2 points power will be deducted by 2

6.25\cdot 10^{-28}

Therefore, Option 4th is correct that is:6.25\cdot 10^{-28}

ioda3 years ago
6 0

(4 x 10^13)^-2

(4)^-2 x (10^-26)

.0625 x 10^-26 = 6.25 x 10 ^ -28

                The underlined expression is the right answer because it follows the rule of scientific notation.

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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

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:

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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.

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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

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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

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<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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