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kati45 [8]
3 years ago
10

If the circumference of a circle is 12.56 ft, what is the length of the diameter? use 3.14 for Ï.

Mathematics
1 answer:
irga5000 [103]3 years ago
5 0
ΠD = C
3.14D = 12.56
D = 12.56/3.14
D =  4

<span>the length of the diameter is 4 ft</span>
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Answer:

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Step-by-step explanation:

The angle JKL is being bisected by line KM. This means that the angle has been split into 2 equal angles. Therefore, one of these angles will be equal to 1/2 of the large angle. So, option B, 1/2(m∠JKL) = m∠JKM, is correct.

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Answer:

Standard Form           Equivalent Form            Extreme Values

y=x^2-6x+17                   (x-3)^2+8                        (3,8)

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y=x^2-16x+60                 (x-8)^2-4                         (8,-4)

Step-by-step explanation:

1) Standard form:

y=x^2-6x+17

Equivalent Form:

Can be found using completing the square method.

y=x^2-6x+17\\y=x^2-2(x)(3)+(3)^2-(3)^2+17\\y=(x-3)^2-9+17\\y=(x-3)^2+8

So, Equivalent form is: (x-3)^2+8

Extreme value:

Extreme values are basically the minimum and maximum value of the function.

Minimum Value will be found by finding derivative of the function:

The derivate is: 2x-6

Now, put the derivate equal to zero: 2x-6 = 0

2x=6\\x=6/3 \\x=3

Maximum value can be found by putting minimum value in the given function:

Put x = 3 and solve:

(3)^2-6(3)+17\\9-18+17\\9-1\\=8\\

So, the extreme values is: (3,8)

2) Standard form:

y=x^2+8x+21

Equivalent Form:

Can be found using completing the square method.

y=x^2+8x+21\\y=x^2+2(x)(4)+(4)^2-(4)^2+21\\y=(x+4)^2-16+21\\y=(x+4)^2+5

So, Equivalent form is: (x+4)^2+5

Extreme value:

Extreme values are basically the minimum and maximum value of the function.

Minimum Value will be found by finding derivative of the function:

The derivate of x^2+8x+21 is: 2x+8

Now, put the derivate equal to zero:

2x+8 = 0\\2x=-8\\x=-8/2 \\x=-4

So, minimum value is: -4

Maximum value can be found by putting minimum value in the given function:

Put x = -4 and solve:

x^2+8x+21\\=(-4)^2+8(-4)+21\\=16-32+21\\=5

So, Maximum value is: 5

So, the extreme values is: (-4,5)

3) Standard form:

y=x^2-16x+60

Equivalent Form:

Can be found using completing the square method.

y=x^2-16x+60\\y=x^2-2(x)(8)+(8)^2-(8)^2+60\\y=(x-8)^2-64+60\\y=(x-8)^2-4

So, Equivalent form is: (x-8)^2-4

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Extreme values are basically the minimum and maximum value of the function.

Minimum Value will be found by finding derivative of the function:

The derivate of x^2-16x+60 is: 2x-16

Now, put the derivate equal to zero:

2x-16 = 0\\2x=16\\x=16/2 \\x=8

So, minimum value is: 8

Maximum value can be found by putting minimum value in the given function:

Put x = 8 and solve:

x^2-16x+60\\=(8)^2-16(8)+60\\=64-128+60\\=-4

So, Maximum value is: -4

So, the extreme values is: (8,-4)

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