All categories

4 years ago

Write a rule to describe the function shown. x y −6 −4 −3 −2 0 0 3 2

Mathematics

1 answer:

Vedmedyk [2.9K]4 years ago

5 0

Answer:

y= 2/3 x

Step-by-step explanation:

First plot the points on a Cartesian plan

You will notice a straight line passing through the origin (0,0)

Find the gradient of the line;

m=change in y/change in x

m=2-0/3-0 =2/3

Find the equation of the line

y-2/x-3 =2/3

3(y-2) = 2(x-3)

3y-6= 2x-6

3y=2x-6+6

3y=2x

y=2/3 x + 0

You might be interested in

Find the mixed number of 18/3??

11Alexandr11 [23.1K]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Which one should I choose

Alona [7]

The answer is B linear

3 0

3 years ago

Read 2 more answers

20. Joaquin tak 4-hours to clean a pool; Sherrie takes 1--hours. Calculate the time difference in hours. Express your answer in

Ber [7]

The time difference is 3 hours

3 0

4 years ago

Rewrite 3/100 as the sum of unit fractions

Bingel [31]

1/100+1/100+1/100 Hope this helped! I remember taking this unit in Elementary. Good luck!

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

4 years ago