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3 years ago

For the function f(x) = (5-6x)^2, fine f^-1

Mathematics

1 answer:

Scorpion4ik [409]3 years ago

8 0

Answer:

f(-1) = 121

Step-by-step explanation:

f(x) = (5-6x)^2 (First we will substitute x for -1)

f(-1) = (5-6(-1))^2 (Then we will combine like terms)

f(-1) = (11)^2 (Then we will square it)

f(-1) = 121 (You are done)

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(-17) x (-46) = find the product

alina1380 [7]

Answer:

782

Step-by-step explanation:

Hope this helps

4 0

3 years ago

Step by Step If 35% of a number is 40, what is the number?

zheka24 [161]

.35x=40
/.35 /.35
X= 40/.35
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4 0

3 years ago

The Cheer Club is making a banner for your school using the design shown. The area of the yellow portion is 1 ft2. The expressio

olga nikolaevna [1]

Answer:

$Blue = \frac{8}{9}$

Step-by-step explanation:

Given

$Yellow = 1ft^2$

$Blue = 8z^{-1$

Required

Evaluate Blue when z = 9

To do this, we simply substitute 9 for z in $Blue = 8z^{-1$

$Blue = 8 * 9^{-1$

Convert indices to fraction

$Blue = 8 * \frac{1}{9}$

$Blue = \frac{8}{9}$

Hence, the blue section has an area of $\frac{8}{9}ft^2$

3 0

3 years ago

Find the equation of the sphere if one of its diameters has endpoints (4, 2, -9) and (6, 6, -3) which has been normalized so tha

Pavel [41]

Answer:

$(x - 5)^2 + (y - 4)^2 + (z - 6)^2 = 14$ .

(Expand to obtain an equivalent expression for the sphere: $x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0$ )

Step-by-step explanation:

Apply the Pythagorean Theorem to find the distance between these two endpoints:

$\begin{aligned}&\text{Distance}\cr &= \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2 + \left(z_2 - z_1\right)^2} \cr &= \sqrt{(6 - 4)^2 + (6 - 2)^2 + ((-3) - (-9))^2 \cr &= \sqrt{56}}\end{aligned}$ .

Since the two endpoints form a diameter of the sphere, the distance between them would be equal to the diameter of the sphere. The radius of a sphere is one-half of its diameter. In this case, that would be equal to:

$\begin{aligned} r &= \frac{1}{2} \, \sqrt{56} \cr &= \sqrt{\left(\frac{1}{2}\right)^2 \times 56} \cr &= \sqrt{\frac{1}{4} \times 56} \cr &= \sqrt{14} \end{aligned}$ .

In a sphere, the midpoint of every diameter would be the center of the sphere. Each component of the midpoint of a segment (such as the diameter in this question) is equal to the arithmetic mean of that component of the two endpoints. In other words, the midpoint of a segment between $\left(x_1, \, y_1, \, z_1\right)$ and $\left(x_2, \, y_2, \, z_2\right)$ would be:

$\displaystyle \left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right)$ .

In this case, the midpoint of the diameter, which is the same as the center of the sphere, would be at:

$\begin{aligned}&\left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right) \cr &= \left(\frac{4 + 6}{2},\, \frac{2 + 6}{2}, \, \frac{(-9) + (-3)}{2}\right) \cr &= (5,\, 4\, -6)\end{aligned}$ .