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3 years ago

Which compound inequality is equivalent to |ax-b|>c for all real numbers a, b, and c, where c>0

Mathematics

2 answers:

Doss [256]3 years ago

8 0

Answer:

Step-by-step explanation:

Keywords

compound inequality, absolute value, equivalent

we have

we know that

To find the compound inequality calculate the two solutions of the absolute value

First solution (case positive)

-------> inequality A

Second solution (case negative)

-------> inequality B

Multiply by all expression

therefore

is equivalent to

and

The answer is

The equivalent compound inequality is

garik1379 [7]3 years ago

6 0

Answer:

D. ax-b<-c or ax-b>c is the correct answer.

Step-by-step explanation:

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Answer:

84+28w

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Step-by-step explanation:

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3 years ago

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A rectangle prism is shown below. What is the volume of the prism?

8_murik_8 [283]

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3 years ago

The sum of two numbers is 40 and the difference is 4. What are the numbers?

Scilla [17]

Answer:

18, 22

Step-by-step explanation:

Half of 40 is 20 right? Then one number has to be 2 less, while the other number is 2 more. So there is a difference of 4.

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3 years ago

4 mi = ___ ft <br> answer please!!

Tpy6a [65]

Answer:

21120

Step-by-step explanation:

1 mile = 5280 feet

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4 0

3 years ago

A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,

algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}$

Where $v_i$ is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is $[0, \pi/2]$ . The critical points of the function are those who make $d'(\theta)=0$ :

$d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}$

$d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...$

The critical value inside the interval is $\pi/4$ .

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft$

The second step is to find the values of the function at the endpoints of the interval:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft$

The biggest value of f is gived by $\pi/4$ , therefore $\pi/4$ is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

4 0

3 years ago