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irga5000 [103]
3 years ago
5

A high school track is shaped as a rectangle with a half circle on either side.

Mathematics
2 answers:
RSB [31]3 years ago
6 0

Answer:

1,207.6 m

Step-by-step explanation:

the big brain hahaha jk i just took the test and got 100%

vesna_86 [32]3 years ago
3 0

Answer:

1,207.6 m

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

we know that

The perimeter of the high school track is equal to the sum of the length of rectangle multiplied by 2 plus the circumference of a circle (remember that the sum of the circumference of two semicircles is equal to the circumference of one circle)

so

P=2L+\pi D

we have

L=96\ m\\D=35\ m

substitute

P=2(96)+(3.14)(35)=302.9\m

Multiply the perimeter by 4 (because Jake plans on running four laps)

so

302.9(4)=1,207.6\ m

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Can someone help with an explanation? This is finding Surface Area.​
Y_Kistochka [10]
The answer should be 95in^2

Explanation:

You’re essentially just finding the area of all the shapes on the triangle and adding them together.

Area of a triangle = 1/2•b•h
Area of a square = l•w or just s2

The area of the triangles given what we have in the diagram can be found with
1/2•5•7 = 17.5

There are 4 triangles so you multiply 17.5 by 4 which gives 70

The base of the triangle (the square) = 25

70+25=95
7 0
3 years ago
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dimaraw [331]

Step-by-step explanation:

f(x)=(x-5)(5x +2)=0

=> x=5 or, x=-2/5

smaller x = -2/5

larger x=5

3 0
3 years ago
Read 2 more answers
Find the product:<br> (-9)(-4)
nadya68 [22]

Answer:

36

Step-by-step explanation:

-9*-4=36

3 0
3 years ago
PLEASE HELP MEE i dont really understand this<br><br> 4(p+1.25)=7 2/5
Reika [66]

Answer:

p=0.6

Step-by-step explanation:

6 0
2 years ago
Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)
USPshnik [31]
See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) +  f(\frac{x_{1} + x_{2} }{2}) + ... +  f(\frac{x_{n-1} + x_{n} }{2})]

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
M_{4} = \frac{1 - 0}{4} [ f(\frac{0 +  \frac{1}{4} }{2}) +  f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) +  f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]
M_{4} = \frac{1}{4} [ f(\frac{1}{8}) +  f(\frac{3}{8}) +  f(\frac{5}{8}) + f(\frac{7}{8})]

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]

M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]

M₄ = 1/4 · (2 - 478/512)
     = 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

6 0
3 years ago
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