All categories

2 years ago

What is the rate of change for the following? (4,5) (2, 15)

Mathematics

2 answers:

Alex2 years ago

6 0

Answer:

-5

Step-by-step explanation:

Given these two points, we can just plug them into the slope formula.

$Slope=\frac{y_2-y_1}{x_2-x_1}$

$\frac{15-5}{2-4}\\\\\frac{10}{-2} \\\\-5$

Evgen [1.6K]2 years ago

3 0

Answer:

-5

Step-by-step explanation:

Rate of Change also slope

Slope = y-y1/x-x1 = tanα = y/x

15-5/2-4 = 10/-2 =-5

You might be interested in

X=2y-1 <br> 3x+4y=11<br> Need help solving

tigry1 [53]

Answer: (-1/5, 2/5)

Step-by-step explanation:

See the photo for explanation and work.

6 0

3 years ago

Over the past month, a hairstylist has had 56 female clients and 24 male clients. What is the ratio of male clients to female cl

irga5000 [103]

Answer:

D.) 7:3

Step-by-step explanation:

First, find the LCM of the numbers, which is 8.

Then divide each number by 8

56÷8=7 24÷8= 3

Then, put them in a ratio

7:3

4 0

2 years ago

Read 2 more answers

Tell whether the ordered pair is a solution to the system.

DerKrebs [107]

Answer:

Step-by-step explanation:

no. substitute -1 for x and -2 for y.

the first one is -1 - 10

this equals -11 not 3

7 0

2 years ago

Read 2 more answers

Find x if y= -2 <br> 4x-6/y=15

worty [1.4K]

Answer:

x=3

Step-by-step explanation:

8 0

2 years ago

A vertical cylinder is leaking water at a rate of 4m3/sec. If the cylinder has a height of 10m and a radius of 2m, at what rate

Lyrx [107]

Answer:

Therefore the rate change of height is $\frac{1}{\pi}$ m/s.

Step-by-step explanation:

Given that a vertical cylinder is leaking water at rate of 4 m³/s.

It means the rate change of volume is 4 m³/s.

$\frac{dV}{dt}=4 \ m^3/s$

The radius of the cylinder remains constant with respect to time, but the height of the water label changes with respect to time.

The height of the cylinder be h(say).

The volume of a cylinder is $V=\pi r^2 h$

$=( \pi \times 2^2\times h)\ m^3$

$\therefore V= 4\pi h$

Differentiating with respect to t.

$\frac{dV}{dt}=4\pi \frac{dh}{dt}$

Putting the value $\frac{dV}{dt}$

$\Rightarrow 4\pi \frac{dh}{dt}=4$

$\Rightarrow \frac{dh}{dt}=\frac{4}{4\pi}$

$\Rightarrow \frac{dh}{dt}=\frac{1}{\pi}$

The rate change of height does not depend on the height.

Therefore the rate change of height is $\frac{1}{\pi}$ m/s.

3 0

2 years ago