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stiks02 [169]
3 years ago
10

Find the inverse function. f (x) = 1/x+2​

Mathematics
1 answer:
zzz [600]3 years ago
4 0

Answer:

f^-1(x)= 1/(x-2)

Step-by-step explanation:

interchange the variables and solve for y.

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What is the solution to the system of equations graphed below y=-2x+4 y=x-5
Tpy6a [65]

Answer:

D. (3, -2)

Step-by-step explanation:

The point where both lines meet is the solution to the system of equations.

3 0
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Find the slope of the line
ikadub [295]

Answer:The answer is negative three

Step-by-step explanation:

8 0
3 years ago
A right triangle has legs that measure 7 cm and 6 cm. What is the length of the hypotenuse in meters?
dalvyx [7]
<h2>Greetings!</h2>

Answer:

\sqrt{85}  or 9.22

Step-by-step explanation:

You need to remember the following equation:

a^{2} + b^{2} = c^{2}

Where a and b is the lengths of the two sides and c is the hypotenuse.

So simply plug these values into this:

6^{2} + 7^{2} = h^{2}

36 + 49 = h^2

36 + 49 = 85

So the hypotenuse is the square root of 85:

\sqrt{85}  = 9.22


<h2>Hope this helps!</h2>
7 0
3 years ago
Determine the graph of the polar equation r =6/2-2cos theta<br> (picture provided)
vredina [299]

Answer:

Choice D is correct

Step-by-step explanation:

The first step is to write the polar equation of the conic section in standard form by dividing both the numerator and the denominator by 2;

r=\frac{3}{1-cos(theta)}

The eccentricity of this conic section is thus 1, the coefficient of cos θ. Thus, this conic section is a parabola since its eccentricity is 1.

The value of the directrix is determined as;

d = k/e = 3/1 = 3

The denominator of the polar equation of this conic section contains (-cos θ) which implies that this parabola opens towards the right and thus the equation of its directrix is;

x = -3

Thus, the polar equation represents a parabola that opens towards the right with a directrix located at  x = -3. Choice D fits this criteria

5 0
3 years ago
3. Which property justifies rewriting
Maksim231197 [3]
B distributive property
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3 years ago
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