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fomenos
2 years ago
9

A plane travels at a speed of 180 mph in still air. Flying with a tailwind, the plane is clocked over a distance of 900 miles. F

lying against a headwind, it takes 2 hours
longer to complete the return trip. What was the wind velocity? (Round your answer to the nearest tenth.)
Mathematics
1 answer:
valina [46]2 years ago
8 0
The answer the answer is appropriately 40
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Three people play a game on each spinner in the previos problem.
arlik [135]

Answer:

Step-by-step explanation:

All other things being equal, spinner A is fine. It is a fair spinner.

Spinner B is not fair. Players 1 and 3 have only 1 number each. Player 2 on the other hand, has 2 numbers that work for him. If player 2 puts in two dollars and players 1 and 3 one dollar each, that should even up the odds. Now you want it just to be fair. So I think player 2 has to put up 2 dollars and players 1 and 3 each put up one. The pot is 4 dollars each time it is played.

Spinner 3 is not fair either. Player one has 2 chances. Player 2 has 3 chances and player 3 has but one chance. There are 6 chances in all. Player 1 should put up 2 dollars to play player 1 should put up 3 dollars and player 1 should put up 1 dollar.

5 0
3 years ago
Is my answer correct? If not please help.
DerKrebs [107]
I think you are correct
4 0
2 years ago
jake noted that the speech of light is approximately 1.1 x 10^9 kh/k the speed of sound is approximately 1.2 x 10^3 kh/k and 11/
Deffense [45]

Answer:

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Step-by-step explanation:

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3 0
2 years ago
A projectile is launched into the air. The function h(t) = –16t2 + 32t + 128 gives the height, h, in feet, of the projectile t s
Zinaida [17]

Answer:

t = 4 seconds

Step-by-step explanation:

The height of the projectile after it is launched is given by the function :

h(t)=-16t^2+32t+128

t is time in seconds

We need to find after how many seconds will the projectile land back on the ground. When it land, h(t)=0

So,

-16t^2+32t+128=0

The above is a quadratic equation. It can be solved by the formula as follows :

t=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}

Here, a = -16, b = 32 and c = 128

t=\dfrac{-32\pm \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=\dfrac{-32+ \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}, \dfrac{-32\- \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=-2\ s\ \text{and}\ 4\ s

Neglecting negative value, the projectile will land after 4 seconds.

4 0
3 years ago
0.2x + 1 = 1.6x + 3.1<br> What value of x satisfies the equation above?
guapka [62]

Answer:

x = -1.5

General Formulas and Concepts:

<u>Pre-Alg</u>

  • Order of Operations: BPEMDAS
  • Equality Properties

Step-by-step explanation:

<u>Step 1: Define equation</u>

0.2x + 1 = 1.6x + 3.1

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Subtract 0.2x on both sides:          1 = 1.4x + 3.1
  2. Subtract 3.1 on both sides:             -2.1 = 1.4x
  3. Divide both sides by 1.4:                 -1.5 = x
  4. Rewrite:                                            x = -1.5

<u>Step 3: Check</u>

<em>Plug in x to verify it's a solution.</em>

  1. Substitution:                    0.2(-1.5) + 1 = 1.6(-1.5) + 3.1
  2. Multiply:                           -0.3 + 1 = -2.4 + 3.1
  3. Add:                                 0.7 = 0.7
5 0
2 years ago
Read 2 more answers
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