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3 years ago

There are 8 elephants in the zoo. Each elephant eats about 2.16 × 105 pounds of vegetation per year.

Mathematics

1 answer:

Andrei [34K]3 years ago

6 0

Answer:

google is key

Step-by-step explanation:

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angle E and angle F are supplementary the measure of angle E is 54 more than the measure of angle F find the measures of each an

Ne4ueva [31]

The measure of angle F is 63° and the measure of angle E is 117°

<u>Step-by-step explanation:</u>

Given that E and F are supplementary angles which means that the sum of the angles is 180°.

Let the measure of angle F=x

given that the measure of angle E is 54 more than F

E=54+x

F+E=180°

x+54+x=180°

2x+54=180°

2x=180-54

=126°

x=126/2=63°

F=63°

E=F+54=63+54=117°

4 0

4 years ago

2a+6=12 what does a equal a=? Please help.

Rashid [163]

Answer:

Step-by-step explanation:

2a+6=12

Step 1: Subtract 6 from both sides.

2a+6=12

-6 -6

2a=6

Step 2: Divide both sides by 2.

2a/2=6/2

a=3

4 0

4 years ago

Read 2 more answers

I need help asap with number four please help me I'll give points

Nat2105 [25]

Idk what it means, sorry and good luck

6 0

4 years ago

Read 2 more answers

PLEASE HELP ME ON THIS

prohojiy [21]

This is in the form mx + b
so,
g(x) = -3x - 5 + 9
-3x + 4
the b is the y-int
y-int = 4

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