Is this graph proportional?

Doug's printing business had a revenue of $6,000 and operating costs of $4,000. How much profit did his business make?

ryzh [129]

$2,000 because operating costs means how much money he needs to keep the business going and revenue is how much he makes

4 0

3 years ago

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Which equation represents the value of x?

shepuryov [24]

Using the Pythagorean theorem you would find X by taking the square root of the base squared (Y^2) subtracted from the hypotenuse squared ( 10^2 = 100)

The equation would be x = √100-y^2

The second answer is the correct one.

7 0

3 years ago

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The transformation of 'f' is represented by 'g'. Write a rule for g. f(x)=\root(3)(x). Then show your check steps.

podryga [215]

Answer:

$g(x)=-2\sqrt[3]x$

or

$g(x) = -2f(x)$

Step-by-step explanation:

Given

$f(x) = \sqrt[3]x$

Required

Write a rule for g(x)

See attachment for grid

From the attachment, we have:

$(x_1,y_1) = (-1,2)$

$(x_2,y_2) = (1,-2)$

We can represent g(x) as:

$g(x) = n * f(x)$

So, we have:

$g(x) = n * \sqrt[3]x$

For:

$(x_1,y_1) = (-1,2)$

$2 = n * \sqrt[3]{-1}$

This gives:

$2 = n * -1$

Solve for n

$n = \frac{2}{-1}$

$n = -2$

To confirm this value of n, we make use of:

$(x_2,y_2) = (1,-2)$

So, we have:

$-2 = n * \sqrt[3]1$

This gives:

$-2 = n * 1$

Solve for n

$n = \frac{-2}{1}$

$n = -2$

Hence:

$g(x) = n * \sqrt[3]x$

$g(x)=-2\sqrt[3]x$

or:

$g(x) = -2f(x)$

4 0

3 years ago

Explain why each of the following integrals is improper. (a) 4 x x − 3 dx 3 Since the integral has an infinite interval of integ

erma4kov [3.2K]

Answer:

a

Since the integral has an infinite discontinuity, it is a Type 2 improper integral

b

Since the integral has an infinite interval of integration, it is a Type 1 improper integral

c

Since the integral has an infinite interval of integration, it is a Type 1 improper integral

d

Since the integral has an infinite discontinuity, it is a Type 2 improper integral

Step-by-step explanation:

Considering a

$\int\limits^4_3 {\frac{x}{x- 3} } \, dx$

Looking at this we that at x = 3 this integral will be infinitely discontinuous

Considering b

$\int\limits^{\infty}_0 {\frac{1}{1 + x^3} } \, dx$

Looking at this integral we see that the interval is between $0 \ and \ \infty$ which means that the integral has an infinite interval of integration , hence it is a Type 1 improper integral

Considering c

$\int\limits^{\infty}_{- \infty} {x^2 e^{-x^2}} \, dx$

Looking at this integral we see that the interval is between $-\infty \ and \ \infty$ which means that the integral has an infinite interval of integration , hence it is a Type 1 improper integral

Considering d

$\int\limits^{\frac{\pi}{4} }_0 {cot (x)} \, dx$

Looking at the integral we see that at x = 0 cot (0) will be infinity hence the integral has an infinite discontinuity , so it is a Type 2 improper integral

7 0

3 years ago

According to the Rational Root Theorem, the following are potential fox) 2x2 +2x 24.roots of -4, -3, 2, 3, 4Which are actual roo

aksik [14]

Correct Answer: First Option

Explanation:

There are two ways to find the actual roots:

a) Either solve the given quadratic equation to find the actual roots

b) Or substitute the value of Possible Rational Roots one by one to find out which satisfies the given equation.

Method a is more convenient and less time consuming, so I'll be solving the given equation by factorization to find its actual roots. To find the actual roots set the given equation equal to zero and solve for x as given below:

$2x^{2} +2x-24=0\\ \\ 2(x^{2} +x-12)=0\\ \\ x^{2} +x-12=0\\ \\ x^{2} +4x-3x-12=0\\ \\ x(x+4)-3(x+4)=0\\ \\ (x-3)(x+4)=0\\ \\ x-3=0, x=3\\ \\ or\\\\x+4=0, x=-4$

This means the actual roots of the given equation are 3 and -4. So first option gives the correct answer.

7 0

4 years ago

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