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Anestetic [448]
2 years ago
12

The slope that passes through the point (-2,-4) and (-3,-8) is

Mathematics
2 answers:
enot [183]2 years ago
7 0

Answer:

y=4x+b (four being the slope)

Step-by-step explanation:

So lets use

ΔY/ΔX

or the y2-y1 over x2-x1

\frac{-8+4}{-3+2}

\frac{-4}{-1}

Therefore, the slope is 4

Vadim26 [7]2 years ago
4 0

Answer:

(-2,-4)

x= -2 and y= -4

(-3,-8)

x= -3 and y= -8

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Simplify this expression
I am Lyosha [343]
Hello there!

Let's put these into simpler terms:
-7s + 3s + 2s 
10 - 8 - 7

-7s + 3s = -4s
-4s + 2s = 2s

10 - 8 = 2
2 - 7 = -5

We are now left with:
2s - 5, which is your simplified expression.

I hope this helps!
8 0
3 years ago
Solve and check the following equation algebraically. 1/2(x + 32) = -10
ANTONII [103]
Solve :
1/2(x + 32) = -10....multiply both sides by 2
x + 32 = -10 * 2
x + 32 = -20
x = -20 - 32
x = - 52 <===

check :
1/2(x + 32) = -10.....when x = -52
1/2(-52 + 32) = -10
1/2(-20) = -10
-20/2 = -10
-10 = -10 (correct)

8 0
3 years ago
Read 2 more answers
Several psychology majors conducted an experiment around campus, asking students to count backwards from 40 to 0 as fast as poss
Dominik [7]

Answer:

Step-by-step explanation:

POINTS:

- Out of every 15 students, 6 were successful in the task.

- The professors' success rate is 140% of the students' success rate.

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5 0
3 years ago
Which of the following shows as a mixed number and as an improper fraction?
Marysya12 [62]

There's no following. You need to post a picture or complete the question.

6 0
3 years ago
Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat
nirvana33 [79]

Answer:

V = \pi r^2 h

For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

\frac{dV}{dt}= \frac{100 L}{min}

Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}

And replacing we got:

\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}

And that represent 0.127 \frac{cm}{min}

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

V = \pi r^2 h

For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

\frac{dV}{dt}= \frac{100 L}{min}

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}

And solving for \frac{dh}{dt} we got:

\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}

We need to convert the rate given into m^3/min and we got:

Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}

And replacing we got:

\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}

And that represent 0.127 \frac{cm}{min}

5 0
3 years ago
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