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dsp73
3 years ago
13

The Ruffs are planning to buy an above-ground swimming pool shaped as a regular octagon. The radius of the octagon is 9 feet. To

the nearest tenth, find the area of the surface of the water in the pool.
Possible answers:

A) 458.2 ft2
B) 553.1 ft2
C) 94.8 ft2
D) 229.1 ft2
Mathematics
2 answers:
const2013 [10]3 years ago
8 0
Hello ,The answer is A
Ainat [17]3 years ago
7 0
Octagon can be split into 8 equal isosceles triangles Area of octagon will be 8X area of each triangle 360/8 = 45 top angle opposite of base is 45 degrees 180(8-2) = 1080, this is sum of interior angles 1080/8 = 135 135/2 = 67.5 bottom 2 angles opposite of radii are 67.5 degrees to find area of triangle we need base length(x) and height(h) use trig ratios h = 9sin(67.5) x = 18cos(67.5) Area = 8*(1/2)*(9sin(67.5))*(18cos(67.5)) Area = 648*sin(67.5)*cos(67.5) Area = 229.1
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8.14÷20 please show work​
saul85 [17]

Answer:

0.407

Step-by-step explanation:

 <u>0. 4 0 7</u>

2 0     /8. 1 4 0

 −<u> 0</u>      

   8 1    

 <u>− 8 0   </u>

     1 4  

  <u> −   0  </u>

     1 4 0

   −<u> 1 4 0</u>

         0

6 0
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
36% of what number is 18?
QveST [7]
Let the no. be x
A/Q
=> 36 × x /100 = 18
=> x = 18 × 100 / 36
=> x = 50
4 0
3 years ago
Read 2 more answers
3.26 round to the nearest tenth
zaharov [31]
It's 3.3 because the 2 is in the tenths place and you look at 6 and say is it over 5? Yes it is so the next number after 2 is 3 so your answer is 3.3
7 0
4 years ago
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gabriel put 6000 in a 2 year CD paying 4% interest, compounded monthly. After 2 years , he withdrew all his money. what was the
shutvik [7]

amount after 2 years   = 6000(1 + (0.04/12))^24  = 6498.86
7 0
3 years ago
Read 2 more answers
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