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Alina [70]
3 years ago
12

A group of learners is trying to identify the vertices of the feasible region from the graph shown below during a Live Classroom

session.
Each learner has a different opinion about one of the points because the intersection of the inequalities for that point are not on the grid lines.

Mathematics
1 answer:
wel3 years ago
3 0

Answer:

what do you want us to answer?

Step-by-step explanation:

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Victor got 24 out of 30 correct on his math test. What percent did Victor get correct?
Snowcat [4.5K]

Answer:

\frac{24}{30} x 100 = 80%

Step-by-step explanation:

8 0
3 years ago
What is y=5x+3 when the x variable is 1
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The answer is 8 y=8 very easyyyy
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3 years ago
Read 2 more answers
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
I need the answer for these.<br> 1a: <br> 1b:<br> 1c:
Fantom [35]

1a) 8 / (1/2) = 16 * 3 = 48


1b) 3sqrt(49) = 3 * 7 = 21


1c) (5+2)(-8) / (-2)^3 -3


(7*-8) / (-8 -3)


-56/-11


56/11

3 0
3 years ago
Barb earns 26% commission on each lab manual she sells. If she sells 1200 manuals at $9.95 each, find her commission.
kvasek [131]
1200 × 9.95 = 11.940 × 26 =310,440 ÷ 100 = 3,104.40 her commission is $ 3,104.40
6 0
3 years ago
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