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allochka39001 [22]

3 years ago

6

6x+9/x-5 x is not equal to 5 find inverse

1 answer:

algol [13]3 years ago

3 0

You have f(x)=(<span>6x+9)/(x-5) and want to find the inverse function? Please note that the parentheses are obligatory here.

1. Replace "f(x)" with y: y = </span>(6x+9)/(x-5)
<span>
2. Interchange the variables x and y: x</span> = (6y+9)/(y-5)

3. Solve this result for y: x(y-5) = (6y+9), or xy-5x = 6y+9. Then:

xy-6y = 9 + 5x, or y(x-6) = 9+5x
9+5x
Then y = --------------
x-6

Here x may not = 6, because division by zero would occur otherwise.

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A right cylinder has a radius of 3 and a height of 12. What is its surface area? O A. 72 units O B. 90 units O c. 45rt units O D

Yuliya22 [10]

Answer:

$A=72\pi\ units^2$

Step-by-step explanation:

Given that,

The radius of the cylinder, r = 3

Height of the cylinder, h = 12

We need to find the surface area of the cylinder. The formula for the surface area of a cylindrical object is given by :

$A=2\pi rh\\\\=2\times \pi\times 3\times 12\\\\=72\pi\ units^2$

So, the surface area of the cylinder is equal to $72\pi\ units^2$ .

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3 years ago

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uranmaximum [27]

Answer:

D

Step-by-step explanation:

The seven should be on the outside because it is on the denominator

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3 years ago

Solve the equation by completing the square. round to the nearest hundredth. x^-4x=5

satela [25.4K]

X^2 - 4x = 5
(x - 2)^2 = 5 + 4
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answer x = -1, 5

7 0

4 years ago

Can someone PLZZZZ HELP!!!!

swat32

Answer:

The solution is:

Part A. $\sqrt{5}^{\frac{7k}{3}})$ which is sqrt(5)^7k/3[/tex]

Part B. k = 18/7

Step-by-step explanation:

Part A.

To solve this part, we're going two use THREE important properties of exponents:

1. $(x^{n})^{m} = x^{nm}$

2. $\frac{x^{n}}{x^{m}} = x^{n-m}$

3. $\sqrt[n]{x^{m}} = x^{\frac{m}{n} }$

Let's work the numerator using the properties 1, 2 and 3:

$(\sqrt{5}^{3} )^{\frac{k}{9} } } = (\sqrt{5}^{3\frac{k}{9}}) = (\sqrt{5}^{\frac{k}{3}})$

Let's work the denominator using the properties 1, 2 and 3:

$(\sqrt{5}^{6} )^{-\frac{k}{3} } } = (\sqrt{5}^{6\frac{k}{3}}) = (\sqrt{5}^(2k))$

Now dividing the numerator by the denominator:

$\sqrt{5}^{\frac{k}{3}-(-2k)})=\sqrt{5}^{\frac{7k}{3}})$

Part B

if $5^{\frac{3}{2} } 5^{\frac{3}{2}} = \sqrt{5}^{\frac{7k}{3}})$

Then:

$5^{3} = 5^{\frac{7k}{6}})$

So $\frac{7k}{6}} = 3$

Solving for k, we have:

k = 18/7

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3 years ago

Write this number in word form 3,388,198

Minchanka [31]

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3 years ago

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