If the area of the region bounded by the curve
and the line
is
Sq units, then the value of
will be
.
<h3>What is area of the region bounded by the curve ?</h3>
An area bounded by two curves is the area under the smaller curve subtracted from the area under the larger curve. This will get you the difference, or the area between the two curves.
Area bounded by the curve
We have,
⇒ 
,
Area of the region
Sq units
Now comparing both given equation to get the intersection between points;

So,
Area bounded by the curve
![\frac{256}{3} =\[ \int_{0}^{4a} \sqrt{4ax} \,dx \]](https://tex.z-dn.net/?f=%5Cfrac%7B256%7D%7B3%7D%20%3D%5C%5B%20%20%5Cint_%7B0%7D%5E%7B4a%7D%20%5Csqrt%7B4ax%7D%20%20%5C%2Cdx%20%5C%5D)
![\frac{256}{3}= \[\sqrt{4a} \int_{0}^{4a} \sqrt{x} \,dx \]](https://tex.z-dn.net/?f=%5Cfrac%7B256%7D%7B3%7D%3D%20%20%20%5C%5B%5Csqrt%7B4a%7D%20%20%5Cint_%7B0%7D%5E%7B4a%7D%20%5Csqrt%7Bx%7D%20%20%5C%2Cdx%20%5C%5D)
![\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{1}{2}+1 } }{\frac{1}{2}+1 }\end{array}\right] _{0}^{4a}](https://tex.z-dn.net/?f=%5Cfrac%7B256%7D%7B3%7D%3D%202%5Csqrt%7Ba%7D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cfrac%7B%28x%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%2B1%20%7D%20%7D%7B%5Cfrac%7B1%7D%7B2%7D%2B1%20%7D%5Cend%7Barray%7D%5Cright%5D%20_%7B0%7D%5E%7B4a%7D)
![\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{3}{2} } }{\frac{3}{2} }\end{array}\right] _{0}^{4a}](https://tex.z-dn.net/?f=%5Cfrac%7B256%7D%7B3%7D%3D%202%5Csqrt%7Ba%7D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cfrac%7B%28x%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%7D%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%5Cend%7Barray%7D%5Cright%5D%20_%7B0%7D%5E%7B4a%7D)
![\frac{256}{3}= 2\sqrt{a} *\frac{2}{3} \left[\begin{array}{ccc}(x)^{\frac{3}{2}\end{array}\right] _{0}^{4a}](https://tex.z-dn.net/?f=%5Cfrac%7B256%7D%7B3%7D%3D%202%5Csqrt%7Ba%7D%20%2A%5Cfrac%7B2%7D%7B3%7D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%28x%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20_%7B0%7D%5E%7B4a%7D)
On applying the limits we get;
![\frac{256}{3}= \frac{4}{3} \sqrt{a} \left[\begin{array}{ccc}(4a)^{\frac{3}{2} \end{array}\right]](https://tex.z-dn.net/?f=%5Cfrac%7B256%7D%7B3%7D%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Csqrt%7Ba%7D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%284a%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%20%5Cend%7Barray%7D%5Cright%5D)



⇒ 

Hence, we can say that if the area of the region bounded by the curve
and the line
is
Sq units, then the value of
will be
.
To know more about Area bounded by the curve click here
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Answer: 110-35=x
answer: 7 minutes
I came here to find the answer to this proble but since no one answered I’ll just put the answer that I got.
Step-by-step explanation:
Answer:
Sam made it -24 versus 24 (29-5=24 not -24)
(Sam failed to flip the inequality)
Step-by-step explanation:
-4x + 5 > 29
Subtract 5 from each side
-4x+5-5 > 29-5
-4x > 24 Sam made it -24 versus 24 (29-5=24 not -24)
Divide by -4 Remember to flip the inequality
-4x/-4 < 24/-4 (Sam failed to flip the inequality)
x < -6
Answer:

Step-by-step explanation:
Whenever we are doing unit conversions, we have to remember to make the first ratio and then each successive ratio should be put <em>in such a way</em> that the top and bottom cancels out with each unit and whatever we want, remains.
A simple example would be if we were to just convert from cm to mm. Suppose we want to convert 5 centimeters to millimeters. We can write the ratios is 2 ways:
First:

Second:

The first one is correct since we want mm, mm should be on top and cm should be on bottom form cm and cm to get canceled.
Now, for our problem, we want mm on top and day on bottom. So, cm and year should cancel. Looking at the first choice, the conversion factors are correct & skimming through, we see that
cm and cm cancels (top and bottom), also year and year cancels (top and bottom), it will leave us with mm on top and days on bottom, <em>which is what we want</em>.
First answer choice is right.
It would take both Hannah and destiny 28 hours to paint the room together