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katrin [286]
3 years ago
11

A plane loses altitude at the rate of 5 meters per second. It begins with an altitude of 8500 meters. The planes altitude is a f

unction of the number of seconds that pass. Write an equation modeling this situation. How much time will pass before the plane lands?
Mathematics
1 answer:
Pavlova-9 [17]3 years ago
5 0

Answer:

28 minutes and 3 seconds

Step-by-step explanation:

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Which expression has a value of 9/10
Luda [366]

Answer:

Me too

Step-by-step explanation:

Is there a good

7 0
3 years ago
What two numbers multiply to -120 and add to -2
USPshnik [31]

Hello!

To answer this question we need to find the factors of -120.

The total set (both negative and postive! ) are the following:

  • 1, -120
  • 2, -60
  • 3, -40
  • 4, -30
  • 5, -24
  • 6, -20
  • 8, -15
  • 10, -12
  • 12, -10
  • 15, -8
  • 20, -6
  • 24, -5
  • 30, -4
  • 40, -3
  • 60, -2
  • 120, -1

After looking over all of these factors, we can see that 10 · -12 is equal to not only -120, but when the multiplication sign is replace by addition, it is equal to -2!

Hope this helps.


5 0
3 years ago
A pipe gallon can pump 1,000 gallons of water in 4 hours, how long would it take for the pipe to fill a 2500 gallon pool?
almond37 [142]
You can divide 1000 by 4 then multiply a number to get 2500

1000/4= 250

250 x 10 = 2500


Answer: 10 hours
8 0
2 years ago
Can anyone solve this? If so tell me and make sure u know what ur doing please
raketka [301]

Answer:

-2

Step-by-step explanation:

To find the slope of a line, you need to find the \frac{rise}{run} between two points. I will be using the points (-3, 2) and (-1, -2).

\frac{rise}{run} = \frac{-2-2}{-1-(-3)}

     = \frac{-2-2}{-1+3}

     = \frac{-4}{2}

     = -2

6 0
2 years ago
If 2tanA=3tanB then prove that,<br>tan(A+B)= 5sin2B/5cos2B-1​
Fed [463]

By definition of tangent,

tan(A + B) = sin(A + B) / cos(A + B)

Using the angle sum identities for sine and cosine,

sin(x + y) = sin(x) cos(y) + cos(x) sin(y)

cos(x + y) = cos(x) cos(y) - sin(x) sin(y)

yields

tan(A + B) = (sin(A) cos(B) + cos(A) sin(B)) / (cos(A) cos(B) - sin(A) sin(B))

Multiplying the right side by 1/(cos(A) cos(B)) uniformly gives

tan(A + B) = (tan(A) + tan(B)) / (1 - tan(A) tan(B))

Since 2 tan(A) = 3 tan(B), it follows that

tan(A + B) = (3/2 tan(B) + tan(B)) / (1 - 3/2 tan²(B))

… = 5 tan(B) / (2 - 3 tan²(B))

Putting everything back in terms of sin and cos gives

tan(A + B) = (5 sin(B)/cos(B)) / (2 - 3 sin²(B)/cos²(B))

Multiplying uniformly by cos²(B) gives

tan(A + B) = 5 sin(B) cos(B) / (2 cos²(B) - 3 sin²(B))

Recall the double angle identities for sin and cos:

sin(2x) = 2 sin(x) cos(x)

cos(2x) = cos²(x) - sin²(x)

and multiplying uniformly by 2, we find that

tan(A + B) = 10 sin(B) cos(B) / (4 cos²(B) - 6 sin²(B))

… = 10 sin(B) cos(B) / (4 (cos²(B) - sin²(B)) - 2 sin²(B))

… = 5 sin(2B) / (4 cos(2B) - 2 sin²(B))

The Pythagorean identity,

cos²(x) + sin²(x) = 1

lets us rewrite the double angle identity for cos as

cos(2x) = 1 - 2 sin²(x)

so it follows that

tan(A + B) = 5 sin(2B) / (4 cos(2B) + 1 - 2 sin²(B) - 1)

… = 5 sin(2B) / (4 cos(2B) + cos(2B) - 1)

… = 5 sin(2B) / (4 cos(2B) - 1)

as required.

5 0
2 years ago
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