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Alinara [238K]
2 years ago
8

F(x)=6x+4. Find the inverse of f(x).

Mathematics
2 answers:
alexdok [17]2 years ago
8 0

Answer:

C

Step-by-step explanation:

f^1(x)= x-6/4

lord [1]2 years ago
4 0

Answer:

c

Step-by-step explanation:

f(x)=6x+4

y=6x+4

x=6y+4

6y=x-4

y=(x-4)/6

f^-1(x)=(x-4)/6

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What is 1/2 divided by 1/3
Artemon [7]

Answer:

\frac{3}{2}

Step-by-step explanation:

\frac{1}{2} ÷ \frac{1}{3}

The reciprocal of  \frac{1}{3} is 3, so finding the reciprocal will replace the division sign with a multiplication sign.

= \frac{1}{2} x 3

= \frac{3}{2} or 1\frac{1}{2} or 1.5

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3 years ago
An oil company fills 1 over 12 of a tank in 1 over 3 hour. At this rate, which expression can be used to determine how long will
Leto [7]
Don't let fractions fool you.

\frac{1}{12} of a tank in \frac{1}{3} an hour is equal to:
1/12 of a tank in 20 minutes.

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Our answer is that it takes 4 hours to fill a tank.

6 0
2 years ago
Raul bought a soft drink and a sandwich for $9.90 . Which is he price of each if the sandwich cost 3.5 times as much as the soft
Radda [10]
B. Sandwich costs $7.70 and the soda costs $2.20
8 0
3 years ago
4. Find the standard from of the equation of a hyperbola whose foci are (-1,2), (5,2) and its vertices are end points of the dia
Ad libitum [116K]

The <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

<h3>How to find the standard equation of a hyperbola</h3>

In this problem we must determine the equation of the hyperbola in its <em>standard</em> form from the coordinates of the foci and a <em>general</em> equation of a circle. Based on the location of the foci, we see that the axis of symmetry of the hyperbola is parallel to the x-axis. Besides, the center of the hyperbola is the midpoint of the line segment with the foci as endpoints:

(h, k) = 0.5 · (- 1, 2) + 0.5 · (5, 2)

(h, k) = (2, 2)

To determine whether it is possible that the vertices are endpoints of the diameter of the circle, we proceed to modify the <em>general</em> equation of the circle into its <em>standard</em> form.

If the vertices of the hyperbola are endpoints of the diameter of the circle, then the center of the circle must be the midpoint of the line segment. By algebra we find that:

x² + y² - 4 · x - 4 · y + 4= 0​

(x² - 4 · x + 4) + (y² - 4 · y + 4) = 4

(x - 2)² + (y - 2)² = 2²

The center of the circle is the midpoint of the line segment. Now we proceed to determine the vertices of the hyperbola:

V₁(x, y) = (0, 2), V₂(x, y) = (4, 2)

And the distance from the center to any of the vertices is 2 (<em>semi-major</em> distance, a) and the semi-minor distance is:

b = √(c² - a²)

b = √(3² - 2²)

b = √5

Therefore, the <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

To learn more on hyperbolae: brainly.com/question/27799190

#SPJ1

5 0
1 year ago
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