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wlad13 [49]
2 years ago
13

A factory produces weighted balls to use for exercise by filling spherical rubber shells of different sizes with the sand like m

aterial. The materials density is 1.5 grams per centimeter. Assuming the shell weighs 10 grams, what should the balls radius so, when full, it weighs 1 kilogram(or 1000 grams)?
Mathematics
2 answers:
anzhelika [568]2 years ago
8 0

Answer:

r=5.4

As mentioned above, the actual \goldE{\text{total quantity}}total quantitystart color #a75a05, start text, t, o, t, a, l, space, q, u, a, n, t, i, t, y, end text, end color #a75a05 is 1000-10=\goldE{990}1000−10=9901000, minus, 10, equals, start color #a75a05, 990, end color #a75a05 grams, since we are trying to find the volume of the air within the shell.

Let's denote the ball's radius as rrr. Then, the \maroonD{\text{volume}}volumestart color #ca337c, start text, v, o, l, u, m, e, end text, end color #ca337c is \maroonD{\dfrac43\pi r^3}  

3

4

​  

πr  

3

start color #ca337c, start fraction, 4, divided by, 3, end fraction, pi, r, cubed, end color #ca337c.

Now we can plug \blueE{\text{density}=1.5}density=1.5start color #0c7f99, start text, d, e, n, s, i, t, y, end text, equals, 1, point, 5, end color #0c7f99, \goldE{\text{total quantity}=990}total quantity=990start color #a75a05, start text, t, o, t, a, l, space, q, u, a, n, t, i, t, y, end text, equals, 990, end color #a75a05, and \maroonD{\text{volume}=\dfrac43\pi r^3}volume=  

3

4

​  

πr  

3

start color #ca337c, start text, v, o, l, u, m, e, end text, equals, start fraction, 4, divided by, 3, end fraction, pi, r, cubed, end color #ca337c in the equation.

nataly862011 [7]2 years ago
6 0

Answer:

5.4

Step-by-step explanation:

Khan-Academy, correct

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Find the value of x when : 2^30/8^9=2^x
artcher [175]

Answer:

x = -3

Step-by-step explanation:

Okay, the first thing to do is get that 8 away from the fraction. How can we do it? When you have 2³ you have 8, is the same thing, so let's do it:

2³⁰/(2³)⁹ = 2^x

When you have a number with the shape (a^x)^y, you can write it as a^(x•y), so:

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liubo4ka [24]

Take the sequence in 1a

The 10th term is 31

The 20th is 61

If you wanted to find these by continuing the series, you'd have to add 3 to the last number in the series, then 3 more, then 3 more, until you reach the 20th term. By this point, you will have added 3 to the first term 19 times. That's where the formula comes from. So here,

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d = 3, how much we're adding each time between one term and the next

Then, to get the 20th term,

4 + (20 - 1) • 3 = 4 + (19 • 3) = 4 + 57 = 61

Answers

The 10th and 20th terms of each sequences are

a. 31; 61

b. 48; 98

c. 47; 97

(in <em>c</em>, you're adding the same <em>d</em> as in the sequence above, but your first term is one unit less)

d. -25; -75

(same thing as before, but now, <em>d</em> is negative)

e. 11.5; 16.5

(with <em>d</em>=1/2 or 0.5)

f. 6+1/2; 8+1/2

Use these to check your answers after applying the formula, but know that I calculated on the fly and didn't check these.

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