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rodikova [14]
3 years ago
12

One circular table has a diameter of 8 ft, and another circular table has a diameter of 12 ft. How much greater is the area of t

he larger table? ​
Mathematics
2 answers:
mezya [45]3 years ago
8 0

Answer:

20π sq ft ≈ 62.83 sq ft

Step-by-step explanation:

Area of table with diameter 8 ft = π*8^2/4 = 16π

Area of table with diameter 12 ft = π*12^2/4 = 36π

Difference between the area of the tables ;

= 36π - 16π

= 20π sq ft ≈ 62.83 sq ft                              ( π = 3.14)

nalin [4]3 years ago
5 0

Answer:

62.83 ft^{2}

Step-by-step explanation:

First, you have to find the area of both using the area formula for a circle:

<em>A = </em>\pir^{2}

Radius is half of the diameter:

<em>A</em>_{1}<em> = </em>(3.1415926535...)( 4^{2} )         <em>A</em>_{2}<em> = </em>(3.14159265...)( 6^{2} )

A_{1} ≈ 50.265...                             A_{2} ≈ 113.097...

How much greater:

A_{2} - A_{1} ≈ 62.83185

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Sonbull [250]

Answer:

-6 < k < 2

Step-by-step explanation:

Given

y = x^2 - 2x

y =kx -4

Required

Possible values of k

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ax^2 + bx + c = 0

Subtract y = x^2 - 2x and y =kx -4

y - y = x^2 - 2x - kx +4

0 = x^2 - 2x - kx +4

Factorize:

0 = x^2 +x(-2 - k) +4

Rewrite as:

x^2 +x(-2 - k) +4=0

Compare the above equation to: ax^2 + bx + c = 0

a = 1

b= -2-k

c =4

For the equation to have two distinct solution, the following must be true:

b^2 - 4ac > 0

So, we have:

(-2-k)^2 -4*1*4>0

(-2-k)^2 -16>0

Expand

4 +4k+k^2-16>0

Rewrite as:

k^2 + 4k - 16 + 4 >0

k^2 + 4k - 12 >0

Expand

k^2 + 6k-2k - 12 >0

Factorize

k(k + 6)-2(k + 6) >0

Factor out k + 6

(k -2)(k + 6) >0

Split:

k -2 > 0 or k + 6> 0

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To make the above inequality true, we set:

k < 2 or k >-6

So, the set of values of k is:

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3 years ago
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Is perpendicular to ? Explain.
olga2289 [7]
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I will do one example for you.

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L2: y = - 3x + 14

L3: y = -x/3 + 14

The slope of a line is the coefficient of the x.

So the slopes are:

L1: slope 3
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L3: slope -1/3

So now multiply the slopes of each pair of lines:

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A

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