All categories

3 years ago

What is the slope of this problem.

Mathematics

1 answer:

Korolek [52]3 years ago

5 0

Answer:

Slope of the line= -1

Step-by-step explanation:

The given line is of equation y= -x

Slope of any line= y/x

= -x/x= -1

You might be interested in

Which of the I-values satisfy the following inequality?

e-lub [12.9K]

Answer:

The answers are options A and B

First solve the inequality

7x < 21

Divide both sides by 7

That's

7x/7 < 21/7

x < 3

The values that satisfy the inequality are

1 and 2 since they are both less than 3

Hope this helps you.

5 0

4 years ago

Read 2 more answers

PLEASE HELP! NEED HELP NOWWWW!!!! PLS! HELPPPP

Sloan [31]

Answer:

$44.50

Step-by-step explanation:

4 0

3 years ago

Question 9 of 9

andrew11 [14]

I’m sure it’s a because when you minus 15.3 and 20 you get 3.15-20

6 0

2 years ago

Write an equation in slope-intercept form where the slope is 5 and y-intercept is 3.

gogolik [260]

Y=mx + b
In this case:
m=slope=5
b=y-intercept=3
y=5x+3

3 0

3 years ago

The rate of change of the volume V of water in a tank with respect to time t is directly proportional to the cubed root of the v

Svetllana [295]

Answer:

The differential equation becomes -

$\frac{dV}{dt} = k\sqrt[3]{V}$ i.e. $\frac{dV}{dt} = kV^{\frac{1}{3} }$

Step-by-step explanation:

Given - The rate of change of the volume V of water in a tank with respect to time t is directly proportional to the cubed root of the volume.

To find - Write a differential equation that describes the relationship.

Proof -

Rate of change of volume V with respect to time t is represented by $\frac{dV}{dt}$

Now,

Given that,

The rate of change of the volume V of water in a tank with respect to time t is directly proportional to the cubed root of the volume.

⇒ $\frac{dV}{dt}$ ∝ $\sqrt[3]{V}$

Now,

We know that, when we have to remove the Proportionality sign , we just put a constant sign.

Let k be any constant.

So,

The differential equation becomes -

$\frac{dV}{dt} = k\sqrt[3]{V}$ i.e. $\frac{dV}{dt} = kV^{\frac{1}{3} }$

6 0

3 years ago