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Ugo [173]
3 years ago
14

MATH PLS HELPP BRAINLIEST U KNO

Mathematics
1 answer:
Viktor [21]3 years ago
7 0

Answer:

Slope is X

Y-intercept is B

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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

3 0
3 years ago
Which unit should be used to measure the amount of water used
Vesna [10]

Answer:

iters and millimets

Step-by-step explanation:

Ur welcome:)

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3 years ago
The stem-and-leaf plot above shows house sale prices over the last week in Tacoma. What was the most
VikaD [51]

Answer:

the answer is 2

Step-by-step explanation:

3 0
2 years ago
Find the power of 1.4 product is 3.<br><br><br>1.64<br><br>2.744<br><br>4.2<br><br>1.96
Natasha2012 [34]

Answer:

the answer would be 4.2

5 0
3 years ago
Read 2 more answers
Question 7 of 10
Ganezh [65]

Answer:

  • Option B

Step-by-step explanation:

Given Equation :

\qquad \sf \dashrightarrow \: 3(4x+3) = 2x - 5(3 - x) + 2

Using distribute property:

\qquad \sf \dashrightarrow \: 12x + 9 = 2x - 15 + 5x + 2

Adding the like terms we get :

\qquad \sf \dashrightarrow \: 12x + 9 = 2x  + 5x  - 15 + 2

\qquad \sf \dashrightarrow \: 12x + 9 = 7x  - 13

Transposing the variables on the right side and constant terms on the left side :

\qquad \sf \dashrightarrow \: 12x  - 7x =   - 13 - 9

\qquad \sf \dashrightarrow \: 5x =   - 22

Dividing both sides by 5 :

\qquad \sf \dashrightarrow \:  \dfrac{5x}{5}  =  \dfrac{ - 22}{5}

\qquad \bf \dashrightarrow \:  x =  \dfrac{ - 22}{5}

3 0
2 years ago
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