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3 years ago

What is 50 + 50? Uduejwnsoeneneof enrkdkdnejelmdkeoqkrjeoskbrnroeodjdneje

Mathematics

2 answers:

Natasha2012 [34]3 years ago

8 0

You add them both is 100

Nookie1986 [14]3 years ago

4 0

Answer:100

Step-by-step explanation:

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Least to greatest 2.3 2 4/5 2.6

Volgvan

2.3 is smaller than 2.6: we can see it just by looking at the digits from left to right.

now, where does 2 $\frac{4}{5}$ stand?

its fraction part is equal to

$\frac{4*2}{5*2}= \frac{8}{10}$

do that means that 2 $\frac{4}{5}$ =2.8

which is bigger than 2.6

so the final order is

2.3 2.6 2 $\frac{4}{5}$

5 0

3 years ago

Jeff had $14.50. She spent $4.35 at the snack bar and $5.25 at the arcade. What is the exact amount of money jeff has left?

Ilia_Sergeevich [38]

Answer:

$4.90

Step-by-step explanation:

14.50-4.35=10.15

10.15-5.25=4.90

7 0

4 years ago

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Tolong dijawab ya kakak

finlep [7]

Answer:

whats the q

Step-by-step explanation:

7 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:

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