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3 years ago

What is the square root of 36x^4

Mathematics

2 answers:

kkurt [141]3 years ago

8 0

He’s right I’ve done the math myself as well

marissa [1.9K]3 years ago

6 0

What is the square root of 36x^4?

√36x^4 =

6x²

(√36=6 and √x^4 = x²)

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Write 12.9 + 8 using words.

Anna35 [415]

Twelve point nine plus eight equals twenty point nine (12.9 + 8 = 20.9)
Hope this helped :)

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3 years ago

FRAME AN EQUATION AND SOLVE IT. NEED HELP ASAP I only have 5 minutes!

lawyer [7]

Answer:

<em>The answer to your question is</em><em> 125</em>

Step-by-step explanation:

<em>The required number is 125.</em>

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2 years ago

Definition of exponential?

lutik1710 [3]

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(of an increase) becoming more and more rapid.

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3 years ago

Read 2 more answers

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

3 years ago

if a right circular cone has a volume of 15 cubic ft and a radius of 3 ft, what is the height of the cone?

Setler79 [48]

Volume of cone=1/3 times height times pir^2
v=15
r=3
15=1/3 hpi3^2
15=1/3hpi9
15=9/3hpi
15=3hpi
divide 3
5=hpi
h=5/pi
pi=aprox 3.14
5/3.14=1.59

1.59 ft

4 0

3 years ago