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Anton [14]
3 years ago
10

I NEED HELP PLEASEEE

Mathematics
2 answers:
Gwar [14]3 years ago
6 0

Answer:

b = \sqrt{333}

Step-by-step explanation:

1. We can use the Pythagorean Theorem to find the missing side, b.

  • 14^2 + b^2 = 23^2

2. (Solving)

Step 1: Simplify both sides of the equation.

  • (14*14)+b^2 =(23*23)
  • 196 + b^2 = 529

Step 2: Subtract 196 from both sides.

  • 196 + b^2 - 196 = 529 - 196
  • b^2 = 333

Step 3: Take square root of both sides.

  • \sqrt{b^2} = \sqrt{333}
  • b = \sqrt{333}

Step 4: Check if solution is correct.

  • 14^2 + \sqrt{333^2} = 23^2
  • (14*14)+(\sqrt{333} *\sqrt{333})=23*23
  • 196 + 333 = 529
  • 529 = 529

Therefore, b = \sqrt{333}.

saw5 [17]3 years ago
4 0

Answer:

Step-by-step explanation:

a^{2} + b^{2} = c^2\\14^2 + b^2 = 23^2\\\\196 + b^2 = 529\\\\b^2 = 333\\b = 18.2482875909

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blondinia [14]

The terms appear to increase by a multiple of -2. The first 7 terms are:

-4 + 8 + -16 + 32 + -64 + 128 + -256. Add these together.

The sum of these numbers is -172.

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How can x-2/3= 5/8 be solved for x in one step
Olenka [21]

Answer:

5/8 + 2/3 = x = 1 7/24

Step-by-step explanation:

Revers the equation. Then add. Well, it's more than one step of course.

5/8 + 2/3 = x

5/8 + 2/3 = 1 7/24

Hope this helps!

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luda_lava [24]

Answer:

P (A and B)=8/9

Events A and B are not dependent.

Step-by-step explanation:

Probability of A AND B means the the probability (favorable outcomes divided by total) of choosing someone who has/did BOTH A and B (not neither or only one of them). 16 did both, so the probability is 16/18 /2/2 = 8/9.

The events are independent (not dependent) because choosing A does not affect choosing B.

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3 years ago
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I need help. Please give me an example
xxTIMURxx [149]

Answer:

9

Step-by-step explanation:

3/729= 9

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=3/9^3= 9


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The after school craft center has 15 boxes of 64 crayons each. In 12 of the boxes , 28 of the crayons have not been used. All th
Korolek [52]

The <em><u>correct answer</u></em> is:

624 crayons.

Explanation:

In 12 of the boxes, 28 crayons have not been used; this leaves 64-28=36 crayons that have been used. 12(36) = 432 crayons have been used in these boxes.

3 full boxes have been used; this is 3(64) = 192 crayons.

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