Complete the equationof the line whose slope is -2 and y-intercept is (0,-8)

Solve for f(3).<br><br> f(x) = 2x – 4

k0ka [10]

Answer: 2

Step-by-step explanation:

All you need to do is plug in 3 for x

2(3)-4

6-4

=2

4 0

3 years ago

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

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8 0

2 years ago

Read 2 more answers

If the length of room is four times the height and breadth is twice of its height and the volume of the room is 512 m^2 find its

lisov135 [29]

$Let \: x \: (x>0) \: be \: the \: height \Rightarrow the \: width, length: 2x,4x \\ Volume:V=hwl = 512 \\ \Leftrightarrow x2x4x = 512 \\ \Leftrightarrow 8{x }^{3} = 512 \\\Leftrightarrow {x}^{3} = 64 \\ \Leftrightarrow x = 4 \\ \Rightarrow The \: height \: is \: 4 \: m, the \: width \: is \: 8 \: m$

8 0

3 years ago

You need to cross a canal and want to determine the distance across the opposite side. Since you're able to take measurements on

MariettaO [177]

Answer:

36.7 ft

Step-by-step explanation:

Measurements for the sides of the canal is given as: 40 ft. and 16 ft.

You solve the above question using Pythagoras Theorem

The distance across the canal is calculated as:

√(40² - 16²)

= √(1344)

= 36.66060556 ft

Approximately = 36.7 ft

Therefore, the distance x across the canal = 36.7 ft

8 0

3 years ago

Find the x-intercept of the line 9× + 3y = -18

ella [17]

Answer:

(-2,0)

Step-by-step explanation:

To find the x intercept, set y =0 and solve for x

9x+3y = -18

9x +0 = -18

9x/9 = -18/9

x = -2

(-2,0)

4 0

3 years ago