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3 years ago

12

Transformations questions! thank you :)

1 answer:

denis-greek [22]3 years ago

4 0

Answer:

Either B or D is the answer

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a cone and a sphere have the same volume. the height of the cone is 96 units. what would be the values for the radius of the con

jolli1 [7]

The volume $V_c$ of a cone with base radius $r_c$ and height $h_c$ is

$V_c = \dfrac{1}{3}\pi r_c^2 h_c$

Similarly, the volume $V_s$ of a sphere with radius $r_s$ is

$V_s = \dfrac{4}{3}\pi r_s^3$

We know that $V_c=V_s$ and that $h_c=96$

So, we can set up the following equation:

$\dfrac{96}{3}\pi r_c^2=\dfrac{4}{3}\pi r_s^3$

We can simplify the common denominator 3, and pi appearing on both sides:

$96r_c^2=4r_s^3$

We can divide both sides by 4:

$24r_c^2=r_s^3$

Without further information, this is all we can say: the cubed radius of the sphere is the same as 24 times the squared radius of the cone.

3 0

4 years ago

Which pair of funtions is not a pair of inverse functions? please help!!

antiseptic1488 [7]

Answer:

$f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Step-by-step explanation:

we know that

To find the inverse of a function, exchange variables x for y and y for x. Then clear the y-variable to get the inverse function.

we will proceed to verify each case to determine the solution of the problem

<u>case A)</u> $f(x)=\frac{x+1}{6} , g(x)=6x-1$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y+1}{6}$

Isolate the variable y

$6x=y+1$

$y=6x-1$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=6x-1$

therefore

f(x) and g(x) are inverse functions

<u>case B)</u> $f(x)=\frac{x-4}{19} , g(x)=19x+4$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y-4}{19}$

Isolate the variable y

$19x=y-4$

$y=19x+4$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=19x+4$

therefore

f(x) and g(x) are inverse functions

<u>case C)</u> $f(x)=x^{5}, g(x)=\sqrt[5]{x}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=y^{5}$

Isolate the variable y

fifth root both members

$y=\sqrt[5]{x}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\sqrt[5]{x}$

therefore

f(x) and g(x) are inverse functions

<u>case D)</u> $f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y}{y+20}$

Isolate the variable y

$x(y+20)=y$

$xy+20x=y$

$y-xy=20x$

$y(1-x)=20x$

$y=20x/(1-x)$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=20x/(1-x)$

$\frac{20x}{1-x}\neq \frac{20x}{x-1}$

therefore

f(x) and g(x) is not a pair of inverse functions

7 0

3 years ago

Read 2 more answers

-4-(-8) divided by 4

Nataly_w [17]

Answer:

1

Step-by-step explanation:

-4 minus -8 is the same as 8 - 4 which is 4 and 4 divided by 4 is 1.

7 0

2 years ago

Read 2 more answers

How many times dose 42 go into 1,984

umka2103 [35]

Here is the answer all you have to do is go 1984 divide by 42 47.23809524

4 0

4 years ago

Solve the equation.<br><br>5/7=10/x+2

skelet666 [1.2K]

Answer:

$x=-\frac{70}{9}$

Step-by-step explanation:

Isolate the variable by doing operations to both sides of the equation.

The work is shown below:

$\frac{5}{7} =\frac{10}{x} +2\\\\\frac{5}{7} - 2=\frac{10}{x} +2 - 2\\\\-\frac{9}{7} = \frac{10}{x}\\\\ -\frac{9}{7}*x = \frac{10}{x}*x\\\\ -\frac{9}{7}x = 10\\\\ -\frac{9}{7}x * (-\frac{7}{9})= 10 * (-\frac{7}{9}) \\\\x=-\frac{70}{9}$

6 0

3 years ago

Read 2 more answers