<img src="https://tex.z-dn.net/?f=%287x%5E%7B2%7D%29%282x%5E%7B4%7D%29%28-5x%29" id="TexFormula1" title="(7x^{2})(2x^{4})(-5x)"

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spayn [35]

2 years ago

alt="(7x^{2})(2x^{4})(-5x)" align="absmiddle" class="latex-formula">

plz help

Mathematics

1 answer:

ivolga24 [154]2 years ago

6 0

Answer:

$-70x^{7}$

Step-by-step explanation:

$(7x^{2})(2x^{4})(-5x)$

$= 7(2)(-5)(x^{2})(x^{4})(x)$

$= -70(x^{7})$

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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

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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3 years ago

What are opposite sides that are parallel

son4ous [18]

Answer: parallel means 2 sides that will never intersect

Step-by-step explanation:

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Susanna has played the piano for s years. Patrick has played the piano for 4 more than twice the number of years that Susanna ha

scoray [572]

S=Years Susanna has played the piano
p=Years Patrick has played the piano

The expression for this question would be s(2)+4=p

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2 years ago

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If x – 9 is a factor of x2 – 5x – 36, what is the other factor? x – 4 x 4 x – 6 x 6

Sergio [31]

Answer:

(x+4)

Step-by-step explanation:

Solve by factoring. Think about what numbers multiply to be -36 (the last term) BUT also subtract/add up to be -5, which is the middle term. The answer to this is 4 and -9. Therefore, the two factors are (x-9)(x+4)

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3 years ago

What is the measure of angle I, m

strojnjashka [21]

Answer:

angle I =110

70 +I = 180 (opp side of parallelogram are equal)

I = 180 - 70

angle I =110

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2 years ago