All categories

3 years ago

What is the graph of linear inequality 2y>x-2

Mathematics

1 answer:

pickupchik [31]3 years ago

3 0

Answer:

Please refer to the attached graph for the graph of linear inequality:

$2y>x-2$

Step-by-step explanation:

Method to draw graph of inequality:

1. First of all replace the inequality sign with equality.

Now the equation is:

$2y=x-2$

Now, find point and plot them.

Let us put x = 0, we get y = -1. So one point is (0, -1)

Let us put y = 0, we get x = 2. So one point is (2,0)

Now, plot both the point and join with straight line on xy coordinate axis.

2. Now, try to find a random point and put in the inequality.

If the point satisfies the inequality shade the region moving from the line towards that point.

For example:

Taking a point (4, 2)

Putting in the inequality:

$8 > 2-2 \\8>0$

So, (4, 2) satisfies the inequality.

Now, shade the region moving from the line towards the point (4,2).

We get the shaded region as shown in the attached figure in the answer area.

You might be interested in

Help me please :))) have a good day also

Ede4ka [16]

1) 2+12a
2) 8-5x
3)12-3z
i think those are the answers :)

8 0

2 years ago

Given I=prt,where I=4.5,p =500,and t=2.5,what is r as a percent?

KIM [24]

PLUG in the numbers
4.5=500 x R x 2.5
divide both sides by 500
0.009=R x 2.5
divide 2.2 to both sides
R=.0036
In percent you have to multiply by 100 which is 0.36%

8 0

3 years ago

Ebony walked at a rate of 3 1/2 miles per hour for 1 1/3 hours. how far did she walk.

Pachacha [2.7K]

About 4.65 miles. Multiply 3 1/2 (3.5) by 1 1/3 (1.33) and thats how you get the answer

7 0

3 years ago

A. Evaluate ∫20 tan 2x sec^2 2x dx using the substitution u = tan 2x.

irakobra [83]

Answer:

The integral is equal to $5\sec^2(2x)+C$ for an arbitrary constant C.

Step-by-step explanation:

a) If $u=\tan(2x)$ then $du=2\sec^2(2x)dx$ so the integral becomes $\int 20\tan(2x)\sec^2(2x)dx=\int 10\tan(2x) (2\sec^2(2x))dx=\int 10udu=\frac{u^2}{2}+C=10(\int udu)=10(\frac{u^2}{2}+C)=5\tan^2(2x)+C$ . (the constant of integration is actually 5C, but this doesn't affect the result when taking derivatives, so we still denote it by C)

b) In this case $u=\sec(2x)$ hence $du=2\tan(2x)\sec(2x)dx$ . We rewrite the integral as $\int 20\tan(2x)\sec^2(2x)dx=\int 10\sec(2x) (2\tan(2x)\sec(2x))dx=\int 10udu=5\frac{u^2}{2}+C=5\sec^2(2x)+C$ .

c) We use the trigonometric identity $\tan(2x)^2+1=\sec(2x)^2$ is part b). The value of the integral is $5\sec^2(2x)+C=5(\tan^2(2x)+1)+C=5\tan^2(2x)+5+C=5\tan^2(2x)+C$ . which coincides with part a)

Note that we just replaced 5+C by C. This is because we are asked for an indefinite integral. Each value of C defines a unique antiderivative, but we are not interested in specific values of C as this integral is the family of all antiderivatives. Part a) and b) don't coincide for specific values of C (they would if we were working with a definite integral), but they do represent the same family of functions.

3 0

3 years ago

Write and test a conjecture about the quotient of a negative number and a positive number. Give at least two examples when testi

4vir4ik [10]

Answer:A negative number divided by a positive number is always negative

Step-by-step explanation:

5 0

3 years ago