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Lubov Fominskaja [6]
3 years ago
8

write a function that represents the situation: A population of 210,000 increases by 12.5% each year​

Mathematics
2 answers:
topjm [15]3 years ago
8 0
Y=2.5x+12,000 i think
Shkiper50 [21]3 years ago
3 0

Answer

y= 12.5x + 210,000

Step-by-step explanation:

This is a linear function because it is increasing constantly by 12.5 percent so it will me written as y=mx+b

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How many shares of stock does Julie Norris hold if her share of the dividend is $6.85?

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3 years ago
the path of a rocket can be modeled by the function h(t) =-16t(t-5) seconds and h is the height in feet.
AnnyKZ [126]
-16 feet x -5 seconds = 80ft/s
4 0
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
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
Paul bought 9 total shirts for a total of 72 tee shirts cost 10 and long sleeve shirts cost 7. how many of each shirt did he buy
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6 long sleeve and 3 short sleeves.
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3 years ago
Read 2 more answers
A regular octagon with side length a cm can be made by cutting off
Ksivusya [100]

Answer:

one side = \frac{a\sqrt{2}}{2}

Step-by-step explanation:

if you draw an octagon on a piece of paper, you can draw a square around it, you should be able to see a diagram of this attached, ignore the 6.

Let's say TP = a

since it's a regular octagon, TP = HT

and using the Pythagoras Theorem,  we know a² + b² = c² and thus:

√(AT² + HA²) = HT

and since AT = HA which we will call x, the equation becomes:

√(2x²) = HT = a

rearrange the equation to solve for x and you get:

2x² = a²

x² = \frac{a^{2} }{2}

x = \frac{a}{\sqrt{2}}

which, if you rationalise the denominator, you get:

x = \frac{a\sqrt{2}}{2}

5 0
3 years ago
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