3 years ago

The inverse of 2x^3+3

Mathematics

2 answers:

Rashid [163]3 years ago

6 0

2/x ^6-3 you do the opposite of everything

valkas [14]3 years ago

6 0

To find the inverse, interchange the variables and solve for y.

$fx^{-1}(x)=\frac{\sqrt[3]{4(x-3)}}{2}$

That answer ^ above should be correct.

You might be interested in

Choose the product. -5p 3(4p 2 + 3p - 1) -20p 5 - 15p 4 + 5p 3 -20p 6 - 15p 4 - 5p 3 -20p 5 + 3p - 1 20p 5 + 15p 4 - 5p 3

Likurg_2 [28]

<span>-5p^3(4p^2 + 3p - 1) = -20p^5 - 15p^4 + 5p^3

answer is

A. first one
</span>-20p^5 - 15p^4 + 5p^3

5 0

2 years ago

Read 2 more answers

If 4 pounds of potatoes costs $6.78. How much does 1 pound of potatoes cost? Show all your work!

baherus [9]

If 4 lbs of potatoes cost 6.78...then 1 lbs of potatoes cost 6.78/4 = 1.695...rounds to 1.70 per lb

3 0

2 years ago

Does the graph represent a function?

Marrrta [24]

Answer:

No, it doesn't pass the vertical line test.

5 0

1 year ago

Substitution would be the simplest method of solution for which system? 5/3 x = 6y and 2x -3y = 8 3x - 2y =18 and 4x + 5y =60 y

LekaFEV [45]

It would be easiest for the last one

3 0

2 years ago

If a cup of coffee has temperature 97Â°C in a room where the ambient air temperature is 25Â°C, then, according to Newton's Law o

FinnZ [79.3K]

Answer:

The average temperature is $T_{a} = 81.95^oC$

Step-by-step explanation:

From the question we are told that

The temperature of the coffee after time t is $T(t) = 25 + 72 e^{[-\frac{t}{45} ]}$

Now the average temperature during the first 22 minutes i.e fro $0 \to 22$ minutes is mathematically evaluated as

$T_{a} = \frac{1}{22-0} \int\limits^{22}_{0} {25 +72 e^{[-\frac{t}{45} ]}} \, dx$

$T_{a} = \frac{1}{22} [25 t + 72 [\frac{e^{[-\frac{t}{45} ]}}{-\frac{1}{45} } ] ] \left| 22} \atop {0}} \right.$

$T_{a} = \frac{1}{22} [25 t - 3240e^{[-\frac{t}{45} ]} ] \left | 45} \atop {{0}} \right.$

$T_{a} = \frac{1}{22} [25 (22) - 3240e^{[-\frac{22}{45} ]} - (- 3240e^{0} )]$

$T_{a} = \frac{1}{22} [550 - 1987.12 + 3240]$

$T_{a} = 81.95^oC$

8 0

2 years ago