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Georgia [21]
3 years ago
15

A cylinder has a height of 16 yards. Its volume is 6,079.04 cubic yards. What is the radius

Mathematics
1 answer:
Katen [24]3 years ago
6 0

Answer:

The radius of the cylinder is: r = 11 yards

Step-by-step explanation:

Given

The height of cylinder = h = 16 yards

Volume of cylinder = V = 6079.04 cubic yards

To determine

The radius of cylinder = r = ?

Using the formula involving volume of the cyliner

V = πr²h

substituting h = 16, V = 6079.04 and π = 3.14

6079.04 = 3.14 (r²) 16

r² = [6079.04] / [(3.14) (16)]

r² = 6079.04 / 50.24

r² = 121

r=\sqrt{121},\:r=-\sqrt{121}

r=11,\:r=-11

As radius can not be negative.

Therefore, the radius of the cylinder is: r = 11 yards

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We have to find x and x is the hypotenuse. We use this formula:

hyp²=side²+side²

So:

hyp²=side²+side²

hyp²= 12²+9²

hyp²= 225

hyp²= √225

hyp= 15

Therefore, the hypotenuse is 15 and x is 15.
HOPE THIS HELPS!
HAVE A GR8 EVENING ;-)

6 0
3 years ago
Need An Answer ASAP ... THANK YOU !!!
lozanna [386]

Answer:

∠EFG = 48°

Step-by-step explanation:

As FH bisects ∠EFG , ∠EFH = ∠HFG .

We know that ∠EFH = (-5x + 89)° . So ∠HFG = ∠EFH = (-5x + 89)°

Also, ∠HFG  + ∠EFH = ∠EFG

=> 2(-5x + 89)° = (61 - x)°

=> -10x + 178 = 61 - x

=> 10x - x = 178 - 61

=> 9x = 117

=> x = 117 / 9 = 13

Putting the value of 'x' in ∠EFG gives :-

(61 - x)° = (61 - 13)° = 48°

8 0
3 years ago
If the radius of the wheel below is 4.2 inches. What is the area of the wheel? Round to the nearest tenth place value.
zepelin [54]

Answer:

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Step-by-step explanation:

6 0
3 years ago
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Factor the polynomial expression x4 + 18x2 + 81
Akimi4 [234]

Answer: (x^2+9)^2

Step-by-step explanation:

Rewrite it in the form a^2+2ab+b^2 , where a=x^2 and b=9.

(x^2)^2+2(x^2)(9)+9^2

Use Square of Sum: (a+b)^2= a^2+2ab+b^2

(x^2+9)^2

4 0
3 years ago
Classify ABC by its sides. Then determine whether it is a right triangle.
m_a_m_a [10]

Answer:

∴Given Δ ABC is not a right-angle triangle

a= AB = √45 = 3√5

b = BC = 12

c = AC = √45 = 3√5

Step-by-step explanation:

Given vertices are A(3,3) and B(6,9)

            AB = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2}  }

            AB = \sqrt{(9-3)^{2}+(6-3)^{2}  } = \sqrt{6^{2}+3^{2}  } =\sqrt{45}

Given vertices are  B(6,9) and C( 6,-3)

       B C = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2}  }

             =  \sqrt{(-3-9)^{2}+(6-6)^{2}  } =\sqrt{12^{2} } = 12

    BC = 12

Given vertices are  A(3,3) and C( 6,-3)

 AC = \sqrt{(6-3)^{2}+(-3-3)^{2}  } = \sqrt{9+36} = \sqrt{45}

AC² = AB²+BC²

45  = 45+144

 45  ≠ 189

∴Given Δ ABC is not a right angle triangle

 

5 0
3 years ago
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