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horsena [70]
3 years ago
7

30 POINTS FOR BRAINLIEST ANSWER!!!!!!!​

Mathematics
1 answer:
skad [1K]3 years ago
6 0
I think b not sure it’s the only one that makes sense
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PLEASE HELP ME ASAP!!
Dmitry_Shevchenko [17]
13 cm+ 13 cm= 26 cm, 26 cm is the answer to Part B.
3 0
3 years ago
Help? with complete solution​ please help me please
serious [3.7K]

9514 1404 393

Answer:

  1. 13 ft
  2. (a) 1 second; (b) t = 0, t = 1/2

Step-by-step explanation:

<h3>1. </h3>

Let w represent the length of the wire. Then the height of attachment is (w-1). The Pythagorean theorem tells us a relevant relation is ...

  5² +(w -1)² = w²

  w² -2w +26 = w² . . . . . . . eliminate parentheses, collect terms

  26 = 2w . . . . . . . . . . . . add 2w

  13 = w . . . . . . . . . . . . divide by 2

The length of the wire is 13 feet.

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<h3>2. </h3>

(a) When h = 0, the equation is ...

  0 = -16t^2 +8t +8

Dividing by -8 puts this into standard form:

  2t^2 -t -1 = 0

Factoring, we get ...

  (2t +1)(t -1) = 0

The positive value of t that makes a factor zero is t = 1.

It will take 1 second for the gymnast to reach the ground.

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(b) When h = 8, the equation is ...

  8 = -16t^2 +8t +8

Subtract 8 and divide by 8 to get ...

  0 = -2t^2 +t

  0 = t(1 -2t) . . . . factor out t

Values of t that make the factors zero are ...

  t = 0

  t = 1/2

The gymnast will be 8 feet above the ground at the start of the dismount, and 1/2 second later.

8 0
2 years ago
How do you solve linear equations using substitution
GuDViN [60]
You substitute the y from the second eqaution to the y in the first equation
4 0
3 years ago
Please help with math
bezimeni [28]

Answer:

First, third, and fourth one would be the answers.

4 0
3 years ago
What is the average of two opposite numbers?
Kryger [21]

Answer:

0

Step-by-step explanation:

So lets imagine x as our number

Let's make an equation:

Average=\frac{x-x}{2} \\Average=\frac{o}{2} \\Average=0

So, This would mean that the average of two opposite numbers is always 0

Hope this helps!

Stay Safe!

8 0
3 years ago
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