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3 years ago

Easy math asap please

Mathematics

2 answers:

Aleks04 [339]3 years ago

7 0

Answer:

Step-by-step explanation:

Firlakuza [10]3 years ago

4 0

The correct Answer is:

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Please help me with this

Lorico [155]

Just apply the Pythagoras Theorem,

LN² = LM² + MN²

LN² = 7² + 8²

LN² = 49 + 64

LN = √113

In short, Your Answer would be: Option D

Hope this helps!

7 0

3 years ago

Read 2 more answers

2/5 of a foot pieces are there in 10 feet?

prohojiy [21]

Answer:

$\frac{a}{25}$ That is the answer if you're finding the exact form...

Step-by-step explanation:

Simplify the expression

5 0

3 years ago

At Healthy Hair, the cost of a child's

Anastasy [175]

Answer:

http://www.elginschools.org/userfiles/179/Classes/7004//userfiles/179/my%20files/module%2011%20review%20key.pdf?id=11326

Step-by-step explanation:

Go There i put the key online with explanation

7 0

3 years ago

Suppose that a police officer measured the speed of cars on a highway for an hour. The average speed of the cars was 72 and the

anygoal [31]

I got 4.14 because i multiplied the deviation which is 4.6 by 0.9 and got that answer

8 0

3 years ago

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