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Akimi4 [234]
3 years ago
10

3^4+4.5= Input whole number

Mathematics
2 answers:
Nina [5.8K]3 years ago
7 0
<h2>86</h2>

Step-by-step explanation:

<h3>(3 × 3 × 3 × 3) + 4.5</h3><h3>81 + 4.5</h3><h3>85.5</h3>

<h2>MARK ME AS BRAINLIST</h2><h2>PLZ FOLLOW ME</h2>
Rasek [7]3 years ago
3 0

Answer:

thats 86 because it's the nearest whole number

Step-by-step explanation:

(3 × 3 × 3 × 3) + 4.5

81 + 4.5

85.5

<h2><em><u>SO THAT MEANS 3^4+4.5 = 85.5 rounded to the nearest whole number is 86</u></em></h2>

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opposite sides are equal

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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.

<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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3 years ago
In parts ​(a) through ​(e)​ below, mark the given statement as True or False. Justify each answer. All vectors are in set of rea
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Answer:

(c)

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Step-by-step explanation:

(a) v  \bullet v = || v ||^2

That is completely correct Remember that if  v = (x_1,x_2,x_3)\\ then

v \bullet v = x_1*x_1+x_2*x_2+x_3*x_3 = x_1^2+x_2^2+x_3^2 = ||v||^2

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"The given statement is true, by definition of length of a vector v, ||v|| = \sqrt{v\bullet v}"

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S_A_V [24]

Answer:

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Step-by-step explanation:

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