All categories

2 years ago

Find the Inverse of 3x^3-1=y

Mathematics

1 answer:

Anika [276]2 years ago

5 0

$y = 3x^{3} - 1$
$x = 3y^{3} - 1$
$3y^{3} = x + 1$
$y^{3} = \frac{x + 1}{3}$

$y = \sqrt[3]{\frac{x + 1}{3}}$

You might be interested in

Find the area. If applicable, leave the answer in simplest radical form. Helpp

inna [77]

Answer:

The area is $28\ m^{2}$

Step-by-step explanation:

I assume that the figure is a square

The area of a square is equal to

$A=b^{2}$

where

b is the length side of the square

In this problem we have

$b=2(\sqrt{7})=2\sqrt{7}\ m$

substitute in the formula

$A=(2\sqrt{7})^{2}$

$A=28\ m^{2}$

6 0

2 years ago

A casino features a game in which a weighted coin is tossed several times. The table shows the probability of each payout amount

kkurt [141]

Answer:

The expected payout of the game is $230.46.

Step-by-step explanation:

The given table is

Payout Amount : $160 $4400 $145000

Probability : 0.146 0.024 0.0007

We need to find the expected payout of the game.

The formula for expected payout is

$\text{Expected payout}=\sum n_iP(x_i)$

where, n is amount and P(x) is probability of that event. The value of n is negative for loss.

Using the above formula we get

$\text{Expected payout}=160\times 0.146+4400\times 0.024+145000\times 0.0007$

$\text{Expected payout}=23.36+105.6+101.5$

$\text{Expected payout}=230.46$

Therefore the expected payout of the game is $230.46.

6 0

2 years ago

The equation y = \large 1\frac{1}{2}x represents the number of cups of dried fruit, y, needed to make x pounds of granola. Deter

Natasha2012 [34]

Answer:

$(1\frac{1}{2},1)$ - False

$(4,6)$ - True

$(18,12)$ -- False

$(0,0)$ -- True

$(2\frac{1}{2},3\frac{3}{4})$ -- True

Step-by-step explanation:

The points are

$(1\frac{1}{2},1)$ , $(4,6)$ , $(18,12)$ , $(0,0)$ and $(2\frac{1}{2},3\frac{3}{4})$ ---- missing from the question

Given

$y = 1\frac{1}{2}x$

Required

Determine if each of the points would be on $y = 1\frac{1}{2}x$

To do this, we simply substitute the value of x and of each point in $y = 1\frac{1}{2}x$ .

(a) $(1\frac{1}{2},1)$

In this case;

$x = 1\frac{1}{2}$ and $y = 1$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 1\frac{1}{2}$

$y = \frac{3}{2} * \frac{3}{2}$

$y = \frac{9}{4}$

$y = 2\frac{1}{4}$

The point $(1\frac{1}{2},1)$ won't be on the graph because the corresponding value of y for $x = 1\frac{1}{2}$ is $y = 2\frac{1}{4}$

(b) $(4,6)$

In this case;

$x = 4$

$y = 6$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 4$

$y = \frac{3}{2} * 4$

$y = \frac{3* 4}{2}$

$y = \frac{12}{2}$

$y = 6$

The point $(4,6)$ would be on the graph because the corresponding value of y for $x = 4$ is $y = 6$

(c) $(18,12)$

In this case:

$x = 18;y = 12$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 18$

$y = \frac{3}{2} * 18$

$y = \frac{3* 18}{2}$

$y = \frac{54}{2}$

$y = 27$

The point $(18,12)$ wouldn't be on the graph because the corresponding value of y for $x = 18$ is $y = 12$

(d) $(0,0)$

In this case;

$x =0; y = 0$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 0$

$y = 0$

The point $(0,0)$ would be on the graph because the corresponding value of y for $x = 0$ is $y = 0$

(e) $(2\frac{1}{2},3\frac{3}{4})$

In this case:

$x = 2\frac{1}{2}$ ; $y = 3\frac{3}{4}$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 2\frac{1}{2}$

$y = \frac{3}{2} * \frac{5}{2}$

$y = \frac{15}{4}$

$y = 3\frac{3}{4}$

The point $(2\frac{1}{2},3\frac{3}{4})$ would be on the graph because the corresponding value of y for $x = 2\frac{1}{2}$ is $y = 3\frac{3}{4}$

3 0

2 years ago

Can someone please help me?

fiasKO [112]

Answer:

?????????????????????????

3 0

2 years ago

How to find lengths from hypotenuse

Phoenix [80]

Can you elaborate on the question??

3 0

2 years ago

Read 2 more answers