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
brainly.com/question/13252576
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Let's start by identify a 30-60-90 triangle
If we compare those two traingles we can stablish the following relation:
![\begin{gathered} a\sqrt[]{3}=6\sqrt[]{3} \\ \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20a%5Csqrt%5B%5D%7B3%7D%3D6%5Csqrt%5B%5D%7B3%7D%20%5C%5C%20%20%5Cend%7Bgathered%7D)
Therefore, a=6
u is the opposite side to the 90° angle, therefore

on the other hand, v is the opposite side to the 30° angles, so
Equation: 2(.25) + p(.75) = 2.75
Answer: 3 pencils
Answer:
I'm pretty sure its A
Step-by-step explanation:
..................
Answer:
(a) k = C + 273.15, F = 1.8(k - 273.15) + 32
(b) k = 296.15
(c) 25 celcius is greater than 300 kelvis.
Step-by-step explanation:
(a) one equation would be to solve for k instead of c, and another related equation could be from kelvin to Farenheit.
C = k - 273.15
C + 273.15 = k
k = C + 273.15
F = 1.8(k - 273.15) + 32
(b) for this one you have to substitute in the previous equation and solve for k.
23 = k - 273.15
k = 23 + 273.15
k = 296.15
(c) you have to convert either kelvins to celcius or celcius to kelvins.
C = k - 273.15
C = 300 - 273.15
C = 26.85
So 25 celcius is greater than 300 kelvis.