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3 years ago

May somebody help me please and thank you

Mathematics

1 answer:

VladimirAG [237]3 years ago

3 0

Answer:

sub to my you tube channel:

Josiah Rowell

Step-by-step explanation:

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Can someone help me with this question?

elena55 [62]

Answer:

238

Step-by-step explanation:

28÷4=7

7x6=42

7x3=21

7x4=28

7x7=49

7x3=21

7x4=28

(28x3)+(21x3)+49+42

84+(21x3)+49+42

84+63+49+42

147+49+42

147+91

238

6 0

2 years ago

Does (0,5) make the equation y = 7x true?

Kay [80]

Answer:

I beleive

Ordered pairs are in the form (x , y). So, my x-coordinate is 0 and my y-coordinate is 0. ... Since, -7 times 0 is equal to 0, I get 0 = 0. This is a true statement, and so (0, 0) is a solution to the equation If its incorrect I'm sorry

Step-by-step explanation:

5 0

3 years ago

Three year ago , Jolene bought $750 worth of stock in a software company. Since then the value of her purchase has been increase

DIA [1.3K]

End of First Year (750/100)x12(3/5)=54 $750+54=804
End of Second Year (804/100)x12(3/5) = 58 $804+58=862
End Of Third Year (862/100)x12(3/5) = 62 $862+62=924

STOCK WORTH NOW = 924

5 0

3 years ago

I get HOW to make a box and whisker plot. What I don't get is WHY you would want to make one. From a practical point of view, wh

ratelena [41]

Answer:

I'd say go to Wikipedia and look it up. It is something that cannot be summarized into just a few words.

Step-by-step explanation:

4 0

4 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

4 years ago