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Lisa [10]
3 years ago
14

Eight expressions are given below. Determine whether each expression is equivalent to 5^10 or not equivalent to 5^10.

Mathematics
1 answer:
KengaRu [80]3 years ago
4 0

Answer:

See Explanation

Step-by-step explanation:

The question is incomplete, as the 8 expressions are not given.

The general explanation is as follows

As complex as an expression may seem, you have to simplify each expression until it cannot be further simplified.

Then you categorize each result, depending on if it equals 5^{10} or not

Take for instance:

5^5 * 5^5

Using law of indices:

5^5 * 5^5 = 5^{5+5

5^5 * 5^5 = 5^{10 ---- equivalent

\frac{5^5 + 5^5}{2 * 5^{-5}}

Factorize the numerator

\frac{5^5 + 5^5}{2 * 5^{-5}} = \frac{5^5(1  + 1)}{2 * 5^{-5}}

\frac{5^5 + 5^5}{2 * 5^{-5}} = \frac{5^5 * 2}{2 * 5^{-5}}

Cancel out 2

\frac{5^5 + 5^5}{2 * 5^{-5}} = \frac{5^5}{5^{-5}}

Apply law of indices

\frac{5^5 + 5^5}{2 * 5^{-5}} = 5^{5--5}

\frac{5^5 + 5^5}{2 * 5^{-5}} = 5^{5+5}

\frac{5^5 + 5^5}{2 * 5^{-5}} = 5^{10 ---equivalent

125 * 78125

Express as exponent

125 * 78125  = 5^3 * 5^7

Apply law of indices

125 * 78125  = 5^{3+7

125 * 78125  = 5^{10 ---- equivalent

5^3 + 5^7

Solve exponents

5^3 + 5^7 = 125 + 78125

5^3 + 5^7 = 78250 ---- not equivalent

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Answer:

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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
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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.

<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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