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mote1985 [20]
3 years ago
6

Sole The Equation 0.75q²=108 55 POINTS

Mathematics
1 answer:
rusak2 [61]3 years ago
5 0
Q= -12

Explain:
Basically we divide both sides by 0.75
Then expand 108 over 0.75 multiplying both by 100 so q cubed= 10800 over 75 when we divide 10800 with 75 we get q cubed=144 then we subtract 144 from each side consider q cubed - 144 rewrite it as q cubed minus 12 cubed the squares can be factored by a cubed and b cubed = ( a - b )(a+b)
So then it’s (q -12) (q+12) = 0 to find solution you gotta solve q - 12 = 0 and q+ 12=0
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What r the blanks 8,570 85.7 8.57
mina [271]

I don't get the question but i'll try...

Well, your 1st blank is going to be 187, and for the second blank...8.850 I think

Your welcome and correct me if i'm wrong.

3 0
3 years ago
(I already did the first half, the Pythagorean theorem part, and the rest is easy I know that for a fact but I don't know how to
Otrada [13]

Answer:

3

Step-by-step explanation:

You'll need to use the Angle Bisector Theorem.

CU / ZU = BC / BZ

From the first part of the problem, you used Pythagorean theorem to find that BZ = 10.

Let's say that CU = x.  That means ZU = 8 − x.  Plugging in values:

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Cross multiply:

10x = 6(8 − x)

Solve:

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4 0
3 years ago
A bag contains five white balls and four black balls. Your goal is to draw two black balls. You draw two balls at random. What i
NARA [144]

Answer:

0.1667 = 16.67% probability that they are both black.

Step-by-step explanation:

The balls are drawn without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this question:

5 + 4 = 9 balls, which means that N = 9

4 are black, which means that k = 4

2 are chosen, which means that n = 2

What is the probability that they are both black?

This is P(X = 2). So

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,9,2,4) = \frac{C_{4,2}*C_{5,0}}{C_{9,2}} = 0.1667

0.1667 = 16.67% probability that they are both black.

8 0
3 years ago
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earnstyle [38]
The answer for this is 60
6 0
3 years ago
For all parts of this question, use the function y - 3 = -2(x - 2)
Elden [556K]

Answer:

Y

5

1

-3

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A step-by-step explanation

5 0
3 years ago
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