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UkoKoshka [18]
3 years ago
9

PLEASE I NEED HELP WITH THIS PROBLEM

Mathematics
2 answers:
LUCKY_DIMON [66]3 years ago
7 0

Answer:

D. -35,-23,-12,58

Step-by-step explanation:

Image a line that goes to the negative, as well as the positives, -35 would be lower on that line

-35          -23       -12        0       12      23          35           58

--------------------------------------------------------------------------------------

Ainat [17]3 years ago
6 0
Negatives would be the least
anything above 0 would normally be the greatest

so d has all the numbers in the correct order
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What is the area of the figure?
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What is the volume of a rectangular prism with a length of 2x, a width of x−2, and a height of x + 2?
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On his first try, Andrew kicked a kickball 21 feet. On his second try, he kicked it 4 times as far. How far did he kick it the s
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The top of a ladder slides down a vertical wall at a rate of 0.25 m/s. At the moment when the bottom of the ladder is 5 m from t
VikaD [51]
<h3>Answer:  13 meters</h3>

==================================================

Explanation:

L = length of ladder

Draw an xy axis. Focus on the upper right quadrant where x > 0 and y > 0

Let the origin (0,0) be the base of the wall.

Let y be the height of the top of the ladder and x be the horizontal distance the base of the ladder is from the base of the wall

Since the ladder is sliding down at a rate of 0.25 m/s, this means dy/dt = -0.25

At x = 5, we are given that dx/dt = 0.6

Use the pythagorean theorem to see that

a^2+b^2 = c^2

x^2 + y^2 = L^2

Now apply the derivative to both sides with respect to t. Don't forget to apply the chain rule while doing so.

d/dt[ x^2 + y^2 ] = d/dt[ L^2 ]

d/dt[ x^2 ] + d/dt[ y^2 ] =  d/dt[ L^2 ]

2x*dx/dt + 2y*dy/dt = 0

2(x*dx/dt + y*dy/dt) = 0

x*dx/dt + y*dy/dt = 0

Note how the right hand side goes to 0 because the derivative of a constant is 0. The ladder length L doesn't change, so neither does L^2.

From here we plug in x = 5, dx/dt = 0.6, dy/dt = -0.25 and we solve for y

x*dx/dt + y*dy/dt = 0

5*0.6 + y*(-0.25) = 0

0.3 - 0.25y = 0

3 = 0.25y

0.25y = 3

y = 3/0.25

y = 12

The top of the ladder is y = 12 meters off the ground and the base is x = 5 meters from the wall.

Therefore,

L^2 = x^2 + y^2

L = sqrt(x^2 + y^2)

L = sqrt(5^2 + 12^2)

L = 13

The ladder is 13 meters long.

6 0
3 years ago
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