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trapecia [35]
3 years ago
14

What is the slope of the line that contains the points (-2, 2) and (3, 4)?

Mathematics
1 answer:
Hitman42 [59]3 years ago
4 0

Answer:

2/5

Step-by-step explanation:

(-2 , 2) = (x1 , x2)

(3 , 4) = (x2 , y2)

slope = y2 - y1/x2 - x1

=4 - 2/3 - (-2)

=2/3+2

=2/5

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What is 15 cm/year converted to inches per month? Round to the nearest tenth.
topjm [15]

Answer:

0.5 inches per month

Step-by-step explanation:

There are nearly 0.3937 inches in 1 cm, so

1 \ cm \approx 0.03937 \ in\\ \\15\ cm\approx 5.9055\ in

There are 12 months in the year, so'

15 centimeters per year is equivalent to

15 \ cm/year\approx \dfrac{5.9055}{12}\ in/month\\\approx 0.492125\ in/month\\\approx 0.5\ in/month\\\\

7 0
4 years ago
Please help me answer this
tekilochka [14]

Answer:

The answer is B

Step-by-step explanation:

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7 0
2 years ago
Heyyy can someone please help me with this ? I was absent when my class went over it.....
Scrat [10]

Answer:

5

Step-by-step explanation:

This is in the form y = mx + b

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b is the y intercept, which is when x = 0 and the line intercepts the y-axis

3 0
3 years ago
What is the solution to the equation (2 – 2) + 22 – 4 = 9?
Alexus [3.1K]

Answer:18

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18≠9

4 0
3 years ago
A ball is thrown into the air and its position is given by h ( t ) = − 6.3 t 2 + 53 t + 24 where h is the height of the ball in
nordsb [41]

Answer:

h'(t) = -12.6 t +53

Now we can set up the derivate equal to 0 and we have:

-12.6 t +53 = 0

And solving for t we got:

t = \frac{53}{12.6}= 4.206

For the second derivate respect the time we got:

h''(t) = -12.6

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

Step-by-step explanation:

For this case we have the following function:

h(t) = -6.3t^2 +53 t+24

In order to maximize this function we need to take the first derivate respect the time and we have:

h'(t) = -12.6 t +53

Now we can set up the derivate equal to 0 and we have:

-12.6 t +53 = 0

And solving for t we got:

t = \frac{53}{12.6}= 4.206

For the second derivate respect the time we got:

h''(t) = -12.6

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

8 0
4 years ago
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