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mojhsa [17]
3 years ago
10

HELP ME!!! I wanna pass please

Mathematics
2 answers:
ale4655 [162]3 years ago
5 0
The answer would be c (w = 12 feet)!
Iteru [2.4K]3 years ago
5 0

2o^2 = w^2+16^2

400 = w^2 +256

400-256 =w^2=256-256

144 =W^2

sqrt 144 =sqrt w^2

12 or -12 =w

w =12 disregard the negative ans C

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Rom4ik [11]

Answer:

My total savings was = $58.33

Step-by-step explanation:

Given:

12% of my savings was = $7.00

To find my total savings.

Solution:

Let the total savings in dollars be = x

12% of the total savings will be  = 12\%\ of\ x

12% of my savings was = $7.00

So, we have:

12\%\ of\ x=7

0.12x=7

Solving for x to get my total savings.

Dividing both sides by 0.12

\frac{0.12x}{0.12}=\frac{7}{0.12}

∴ x=58.33

Thus, my total savings was = $58.33

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3 years ago
PLS HELP WILL GIVE BRAINLIEST AND 20 POINTS TY!
brilliants [131]

Answer: It is B)

Step-by-step explanation: It was 8.66 something and that rounded to the nearest tenth is 8.7

7 0
2 years ago
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Use the expression 14X + 28Y + 21.. what is the greatest common factor
12345 [234]

Answer:

1

Step-by-step explanation:

None of these numbers are alike terms so the only number left is 1.

4 0
2 years ago
A recent survey by the cancer society has shown that the probability that someone is a smoker is P(S)=0.29. They have also deter
GuDViN [60]

<u>Answer:</u>

The correct answer option is P (S∩LC) = 0.16.

<u>Step-by-step explanation:</u>

It is known that the probability if someone is a smoker is P(S)=0.29 and the probability that someone has lung cancer, given that they are also smoker is P(LC|S)=0.552.

So using the above information, we are to find the probability hat a random person is a smoker and has lung cancer P(S∩LC).

P (LC|S) = P (S∩LC) / P (S)

Substituting the given values to get:

0.552 = P(S∩LC) / 0.29

P (S∩LC) = 0.552 × 0.29 = 0.16

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3 years ago
What is the constant of proportionality for the equation y=1 1/2x?​
Dimas [21]

Answer:

<h3>hope it helps you see the attachment for further information... </h3>

<h3>regards..... </h3>

<h3>_addy_✨✨</h3>

6 0
2 years ago
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