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gladu [14]
2 years ago
11

I need help answering this question as soon as possible. i don’t understand it

Mathematics
1 answer:
earnstyle [38]2 years ago
3 0

Answer:

B)

Step-by-step explanation:

Given f(x) = ( 7 - 8x )²

let f(x) be y,

y =  ( 7 - 8x )²  .............to find inverse we have to make x the subject.

±√y = 7 - 8x

-8x = √y  - 7

x = (±√y  - 7 ) / -8

Then,

Inverse of ( 7 - 8x )² is  -\frac{\sqrt{x}-7}{8},\:-\frac{-\sqrt{x}-7}{8} and is a  function.

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Find an equation of the plane that is parallel to the xz-plane and is located 11 units to the right of the xz-plane in standard
mina [271]

Answer:

y = 11

Step-by-step explanation:

Since our plane is parallel to the xz plane, the x and z are independent of each other, i.e. they could be anything. So xz plane has an equation of y = 0. So if this plane is located 11 units to the right of the xz plane then its equation must be y = 11.

8 0
3 years ago
Given a 30-60-90 triangle with a long leg of 9 inches, determine the length of the hypotenuse
lianna [129]

A Quick Guide to the 30-60-90 Degree Triangle

The 30-60-90 degree triangle is in the shape of half an equilateral triangle, cut straight down the middle along its altitude. It has angles of 30°, 60°, and 90°. In any 30-60-90 triangle, you see the following: The shortest leg is across from the 30-degree angle, the length of the hypotenuse is always double the length of the shortest leg, you can find the long leg by multiplying the short leg by the square root of 3.

Note: The hypotenuse is the longest side in a right triangle, which is different from the long leg. The long leg is the leg opposite the 60-degree angle.

Two of the most common right triangles are 30-60-90 and the 45-45-90 degree triangles. All 30-60-90 triangles, have sides with the same basic ratio. If you look at the 30–60–90-degree triangle in radians, it translates to the following:

30, 60, and 90 degrees expressed in radians.

The figure illustrates the ratio of the sides for the 30-60-90-degree triangle.

A 30-60-90-degree right triangle.

A 30-60-90-degree right triangle.

If you know one side of a 30-60-90 triangle, you can find the other two by using shortcuts. Here are the three situations you come across when doing these calculations:

Type 1: You know the short leg (the side across from the 30-degree angle). Double its length to find the hypotenuse. You can multiply the short side by the square root of 3 to find the long leg.

Type 2: You know the hypotenuse. Divide the hypotenuse by 2 to find the short side. Multiply this answer by the square root of 3 to find the long leg.

Type 3: You know the long leg (the side across from the 60-degree angle). Divide this side by the square root of 3 to find the short side. Double that figure to find the hypotenuse.

Finding the other sides of a 30-60-90 triangle when you know the hypotenuse.

Finding the other sides of a 30-60-90 triangle when you know the hypotenuse.

In the triangle TRI in this figure, the hypotenuse is 14 inches long; how long are the other sides?

Because you have the hypotenuse TR = 14, you can divide by 2 to get the short side: RI = 7. Now you multiply this length by the square root of 3 to get the long side:

The long side of a 30-60-90-degree triangle.

6 0
3 years ago
Which figure must be a square?
Maslowich
I think the answer is A

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7 0
4 years ago
Determine the discount between the point(-4,2) and the line 4y=3×+6​
frozen [14]

Answer:

d = 14/5

Step-by-step explanation:

The point (-4,2) means that;

At x = -4, y = 2

Now general form of a linear equation is;

Ax + By + C = 0

We are given;

4y = 3x + 6​

Rearranging to the form of a linear equation gives;

3x - 4y + 6 = 0

Thus, A = 3, B = -4 and C = 6

Thus, at point (-4,2), distance between them is;

d = (3(-4) - 4(2) + 6)/√(3² + (-4)²)

d = -14/5

We will take the absolute value.

Thus; d = 14/5

6 0
3 years ago
Let f(y) = y^4 - 3y^3 + y - 3 and g(y) = y^3 + 7y^2 - 2. Find f(y) + g(y). Write your answer as a polynomial with terms of decre
Anarel [89]
<h3>Answer:   y^4 - 2y^3 + 7y^2 + y - 5</h3>

Work Shown:

h(y) = f(y) + g(y)

h(y) = (y^4 - 3y^3 + y - 3) + (y^3 + 7y^2 - 2)  

h(y) = y^4 - 3y^3 + y - 3 + y^3 + 7y^2 - 2  

h(y) = y^4 + (-3y^3+y^3) + 7y^2 + y + (-3 - 2)

h(y) = y^4 - 2y^3 + 7y^2 + y - 5

This is a fourth degree polynomial (aka quartic).

5 0
2 years ago
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