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ad-work [718]
2 years ago
8

Please help need urgent

Mathematics
1 answer:
aleksandr82 [10.1K]2 years ago
3 0

Convert the cm to feet for the base and the height:

12/2 = 6 x 3 = 18 feet

9/2 = 4.5 x 3 = 13.5 feet


area = 1/2 x base x height

Area = 1/2 x 18 x 13.5

Area = 121.5 square feet


answer: 121.5 square feet

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sveta [45]
Probably d? ........
8 0
3 years ago
The equation below describes a proportional relationship between x and y. What is the constant of proportionality? Y=4/7x
FromTheMoon [43]

Answer:

the given equation is y=4/7x

reciprocating both sides you get y/x =7/4

now x and y are directly porportional to each other so

x=ky where k is the constant

or x/y=k ......................(2)

equating 1 and 2

k=7/4

Step-by-step explanation:

4 0
3 years ago
4s*2+7s+2=0 Quadratic formula
natita [175]

Answer:

z=−215z=-215

Step-by-step explanation:

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6 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.

<h2>Why?</h2>

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
the first grade class has a hamster and a gerbil. the hamster,h, weighs 3 times as much as the gerbil
Serhud [2]

Answer:

The hamster weighs 27 ounces.

Step-by-step explanation:

Since the gerbil weighs 9 ounces and the hamster is 3 times that you must multiply the two numbers together. 9x3=27, so therefore, the hamster weighs 27 ounces.

4 0
4 years ago
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