All categories

3 years ago

Whats 3 square root 64

2 answers:

6 0

$\sqrt[3]{64}$ means you need to find a number that you can multiply by itself 3 times to get 64. In this case, your answer is 4.

4*4*4=64.

Don't forget to rate this answer the Brainliest!

Makovka662 [10]3 years ago

5 0

The answer to your question is 24

You might be interested in

– 42 – 54 + 96 = what is the answer

mart [117]

Answer:

Answer:the answer is zero 

Answer:the answer is zero Step-by-step explanation:

Answer:the answer is zero Step-by-step explanation:when we add -42 and -54 the answer we get is -96. -96and +96 cut each other and then the answer is zero 

 $- 42 - 54 + 96 = 0$ 

6 0

2 years ago

There are 10 fishes 2 drown 1 runs away and 5 died how many fishes are left?

dezoksy [38]

There are still 10 fish but only 5 are alive because fish don't drown

3 0

3 years ago

Read 2 more answers

Select the multiplication equation that could represent the following question: how many 3/8s are in 5/4

muminat

Answer:

The answer is: 3 1/3.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A factory is going to hire 33 new employees. There are 79 applicants. Which

dedylja [7]

Answer:

33:79

Step-by-step explanation:

We are told that a factory is going to hire 33 new employees.

and the total applicants is 79

In this problem, we want to represent the situation as a ratio of new hires to applicants

Since the question says new hires to applicants

Hence, the ratio will be

33 to 79

33/79, we can simplify this as

33:79

5 0

3 years ago

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.

grandymaker [24]

$(2x+1)^{\cot x}=\exp\left(\ln(2x+1)^{\cot x}\right)=\exp\left(\cot x\ln(2x+1)\right)=\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)$

where $\exp(x)\equiv e^x$ .

By continuity of $e^x$ , you have

$\displaystyle\lim_{x\to0^+}\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)=\exp\left(\lim_{x\to0^+}\dfrac{\ln(2x+1)}{\tan x}\right)$

As $x\to0^+$ in the numerator, you approach $\ln1=0$ ; in the denominator, you approach $\tan0=0$ . So you have an indeterminate form $\dfrac00$ . Provided the limit indeed exists, L'Hopital's rule can be used.

$\displaystyle\exp\left(\lim_{x\to0^+}\dfrac{\ln(2x+1)}{\tan x}\right)=\exp\left(\lim_{x\to0^+}\dfrac{\frac2{2x+1}}{\sec^2x}\right)$

Now the numerator approaches $\dfrac21=2$ , while the denominator approaches $\sec^20=1$ , suggesting the limit above is 2. This means

$\displaystyle\lim_{x\to0^+}(2x+1)^{\cot x}=\exp(2)=e^2$

7 0

3 years ago