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Viktor [21]
3 years ago
12

Pls help me asap pls is all im asking for people this is for a final nd like ye im stressin.

Mathematics
1 answer:
sergejj [24]3 years ago
4 0
It wouldn’t let me cancel to answer this but i do not know the answer and i had to type something to exit it. i’m sorry
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If the area of the region bounded by the curve y^2 =4ax and the line x= 4a is 256/3 Sq units, using integration find the value o
almond37 [142]

If the area of the region bounded by the curve y^2 =4ax  and the line x= 4a  is \frac{256}{3} Sq units, then the value of  a will be 2 .

<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  =\int\limits^a_b {x} \, dx

We have,

y^2 =4ax  

⇒ y=\sqrt{4ax}

x= 4a,

Area of the region  =\frac{256}{3}  Sq units

Now comparing both given equation to get the intersection between points;

y^2=16a^2

y=4a

So,

Area bounded by the curve  =   \[ \int_{0}^{4a} y \,dx \] ​

 \frac{256}{3} =\[  \int_{0}^{4a} \sqrt{4ax}  \,dx \]

\frac{256}{3}=   \[\sqrt{4a}  \int_{0}^{4a} \sqrt{x}  \,dx \]

 \frac{256}{3}=   \[2\sqrt{a}  \int_{0}^{4a} x^{\frac{1}{2} }   \,dx \]                                            

\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}

\frac{256}{3}= 2\sqrt{a}  \left[\begin{array}{ccc}\frac{(x)^{\frac{3}{2} } }{\frac{3}{2} }\end{array}\right] _{0}^{4a}

\frac{256}{3}= 2\sqrt{a} *\frac{2}{3}  \left[\begin{array}{ccc}(x)^{\frac{3}{2}\end{array}\right] _{0}^{4a}

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]

\frac{256}{3}= \frac{4}{3} \sqrt{a} *\sqrt{(4a)^{3} }

\frac{256}{3}= \frac{4}{3} \sqrt{a} *  8 *a^{2}   \sqrt{a}

\frac{256}{3}= \frac{4}{3} *  8 *a^{3}

⇒ a^{3} =8

a=2

Hence, we can say that if the area of the region bounded by the curve y^2 =4ax  and the line x= 4a  is \frac{256}{3} Sq units, then the value of  a will be 2 .

To know more about Area bounded by the curve click here

brainly.com/question/13252576

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Solve the equation for theta (0°≤theta&lt;360°)<br><br>2 sin theta = 1​
Oduvanchick [21]

Answer:

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Step-by-step explanation:

To figure out theta is using a bit of simple trigonometry.

2sin(theta) = 1

= sin(theta) = 0.5

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x = arcsin(0.5)

x = 30

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3 1/8 or 3.125 or 25/8
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Help me please, I don't really get this at all.
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Answer:

b.

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Seats to minutes = 2 to 11:

2/11 = s/m.

When  s = 16:

2/11 = 16/m

m = 11*16 / 2

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= 88  minutes.

When s =  19:

2/11 = 19 / m

m = 11*19 / 2

m  = 209/2

= 104.5 minutes.

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A population grew from 100,000 in 1990 to 124.000 in 2000. What was the percent of increase?
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