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-Dominant- [34]

3 years ago

11

Scott's company makes solid balls out of scrap metal for various industrial uses. For one project, he must make copper balls tha

t
have a radius of 6 in. If copper costs $0.75 per in, how much will the copper cost to make one ball?
Use 3.14 for I, and do not round your answer.

1 answer:

Phantasy [73]3 years ago

5 0

Explanation is in a file

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Explain one relationship between capacity and fluid ounces

kvv77 [185]

Fluid ounces are a unit of capacity

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3 years ago

PLZZZZZZ IL do anything give brainilest 5 star i'll like My grade depends on this PLEASE I NEED THIS TO BE CORRECT AND SHOW YOUR

sesenic [268]

Answer:

$195.00

Step-by-step explanation:

perimeter is 54 feet

area is 152 $ft^{2}$

54' * $1.50 = $81.00

152/8 = 19 bags of mulch needed

19*$6.00= $114.00

$195.00 you get away from the store for under 200 :P

P.S. since you said you'll do anything, send me $1,000,000.00 dollars :D

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3 years ago

What is greater 5.3 or 5.30

Simora [160]

5.30 because 3 is smaller than 30

7 0

3 years ago

Read 2 more answers

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Andre45 [30]

Yo sup??

The given points are (1,1),(1,3) and (9,2)

equation of circle is

8x²+8y²+ax+by=0

let us plug in (1,1) in the equation we get

8+8+a+b=0

a+b=-16

let us now plug in (1,3) in the equation

8+8*3²+a+3b=0

a+3b=-80

let us. now solve the two equation in terms of a and b

we get

2b=--64

b=-32

and

a=-16+32

=16

therefore the equation is

8x²+8y²+16x-32y=0

Hope this helps

6 0

3 years ago

If a = √3-√11 and b = 1 /a, then find a² - b²

Serga [27]

If $b=\frac1a$ , then by rationalizing the denominator we can rewrite

$b = \dfrac1{\sqrt3-\sqrt{11}} \times \dfrac{\sqrt3+\sqrt{11}}{\sqrt3+\sqrt{11}} = \dfrac{\sqrt3+\sqrt{11}}{\left(\sqrt3\right)^2-\left(\sqrt{11}\right)^2} = -\dfrac{\sqrt3+\sqrt{11}}8$

Now,

$a^2 - b^2 = (a-b) (a+b)$

and

$a - b = \sqrt3 - \sqrt{11} + \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{9\sqrt3 - 7\sqrt{11}}8$

$a + b = \sqrt3 - \sqrt{11} - \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{7\sqrt3 - 9\sqrt{11}}8$

$\implies a^2 - b^2 = \dfrac{\left(9\sqrt3 - 7\sqrt{11}\right) \left(7\sqrt3 - 9\sqrt{11}\right)}{64} = \boxed{\dfrac{441 - 65\sqrt{33}}{32}}$

5 0

2 years ago