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masha68 [24]
3 years ago
7

Compute the compositions H ∘ J and J ∘ H to determine whether or not J and H are inverses. (Simplify your answers completely.) H

and J are both defined from ℝ − {1} to ℝ − {1} by the formula H(x) = x + 1 x − 1 , J(x) = x + 1 x − 1 for each x ∈ ℝ − {1}
Mathematics
1 answer:
sdas [7]3 years ago
3 0

Answer:

Remember, if H(x) and J(x) are functions, then (H\circ J)(x)=H(J(x)). And G(x) is the inverse function of H(x) if (G\circ H)(x)=(H\circ G)(x)=x

With  H(x)=\frac{x+1}{x-1},\; J(x)=\frac{x+1}{x-1}. Since H and J are the same function, then

(H\circ J)(x)=(J\circ H)(x)=J(\frac{x+1}{x-1})=\\=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}=\frac{\frac{x+1+x-1}{x-1}}{\frac{x+1-x+1}{x-1}}=\frac{2x(x-1)}{2(x-1)}=x.

Since (J\circ H)(x)=(H\circ J)(x)=x, then H and J are inverses.

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Rewrite Linear equation in slope intercept form: 12x-3y=18
shutvik [7]

Answer:

y = 4x - 6

Step-by-step explanation:

12x - 3y = 18 is in standard form. We need to convert it to slope-intercept form, which looks like y = mx + b, with m being the slope and b being the y-intercept.

We know that 12x has to do with our slope, and 18 is our y-intercept. We need to put those on the right side of the equation so that we can isolate y on the left side:

12x - 3y = 18

Subtract 12x from both sides.

-3y = -12x + 18

Now, we can divide both sides by -3 to get y by itself.

y = 4x - 6

And that looks like slope-intercept form (y = mx + b).

Hopefully this was helpful! If you have more questions, let me know. :)

7 0
3 years ago
36 2/5 x what = 109 1/5
tamaranim1 [39]
I am not sure what the answer is but covert them into a decimal and then divide it
6 0
4 years ago
Read 2 more answers
Determine whether the relation below is also a function.
Bess [88]

Answer:

B. Not a function

Step-by-step explanation:

3 0
3 years ago
(10.02)
ollegr [7]
<h2>Hello!</h2>

The answer is:

C. Cosine is negative in Quadrant III

<h2>Why?</h2>

Let's discard each given option in order to find the correct:

A. Tangent is negative in Quadrant I: It's false, all functions are positive in Quadrant I (0° to 90°).

B. Sine is negative in Quadrant II: It's false, sine is negative in positive in Quadrant II. Sine function is always positive coming from 90° to 180°.

C. Cosine is negative in Quadrant III. It's true, cosine and sine functions are negative in Quadrant III (180° to 270°), meaning that only tangent and cotangent functions will be positive in Quadrant III.

D. Sine is positive in Quadrant IV: It's false, sine is negative in Quadrant IV. Only cosine and secant functions are positive in Quadrant IV (270° to 360°)

Have a nice day!

6 0
4 years ago
What is the point-slope form of a line with slope -5 that contains the point<br> a<br> (2,-1)?
gulaghasi [49]

The point-slope form the line which has the value of slope -5 and contains a point as A(2,-1) is (y+1)=-5(x-2).

<h3>What is point slope form?</h3>

The point slope  form of a line is the expression of line which has a specified slope and passes through a point.

The point slope form is givne as,

(y-y₁)=m(x-x₁)

Here, m is the slope of the line, x₁ is the x coordinate of the point by which line passes and y₁ is the y coordinate of the same point.

The slope of a line is -5. This line contains the point A (2,-1). Thus, the point slope form is,

(y-y₁)=m(x-x₁)

(y-(-1))=-5(x-2)

(y+1)=-5(x-2)

Thus, the point-slope form the line which has the value of slope -5 and contains a point as A(2,-1) is (y+1)=-5(x-2).

Learn more about the point slope form here;

brainly.com/question/6497976

#SPJ1

8 0
2 years ago
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