All categories

3 years ago

3/8x−1/3x+6=x find x

Mathematics

1 answer:

Elanso [62]3 years ago

3 0

Step-by-step explanation:

3/8x-1/3x+6=x

(3(3x)-1(8x))/24=x-6

(9x-8x)=24(x-6)

x=24x-24×6

x=24x-144

144=24x-x

23x=144

x=144/23

You might be interested in

Need help for math due tonight!!

Len [333]

Answer:

Use the pythagorian theorm

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

(5.2x10^-1)(1.5x10^-5)

VLD [36.1K]

Answer:

(10^-6) (5.2x1.5 )=10^-6 x 7.8 = 7.8 x10^-6

Step-by-step explanation:

0.000001 x7.8=0.0000078

6 0

3 years ago

Which part of this graph shows a nonlinear relationship?

8_murik_8 [283]

Answer:

D. 4 is the non linear area of the graph.

Step-by-step explanation:

Non linear is when there is not a exact increase or decrease, also known as a equal change over time.

7 0

3 years ago

Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified l

Sloan [31]

Answer:

The integral of the volume is:

$V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy$

The result is: $V = 78.97731$

Step-by-step explanation:

Given

Curve: $x^2 + 4y^2 = 4$

About line $x = 2$ --- Missing information

Required

Set up an integral for the volume

$x^2 + 4y^2 = 4$

Make x^2 the subject

$x^2 = 4 - 4y^2$

Square both sides

$x = \sqrt{(4 - 4y^2)$

Factor out 4

$x = \sqrt{4(1 - y^2)$

Split

$x = \sqrt{4} * \sqrt{(1 - y^2)$

$x = \±2 * \sqrt{(1 - y^2)$

$x = \±2 \sqrt{(1 - y^2)$

Split

$x_1 = -2 \sqrt{(1 - y^2)}\ and\ x_2 = 2 \sqrt{(1 - y^2)}$

Rotate about x = 2 implies that:

$r = 2 - x$

So:

$r_1 = 2 - (-2 \sqrt{(1 - y^2)})$

$r_1 = 2 +2 \sqrt{(1 - y^2)}$

$r_2 = 2 - 2 \sqrt{(1 - y^2)}$

Using washer method along the y-axis i.e. integral from 0 to 1.

We have:

$V = 2\pi\int\limits^1_0 {(r_1^2 - r_2^2)} \, dy$

Substitute values for r1 and r2

$V = 2\pi\int\limits^1_0 {(( 2 +2 \sqrt{(1 - y^2)})^2 - ( 2 -2 \sqrt{(1 - y^2)})^2)} \, dy$

Evaluate the squares

$V = 2\pi\int\limits^1_0 {(4 +8 \sqrt{(1 - y^2)} + 4(1 - y^2)) - (4 -8 \sqrt{(1 - y^2)} + 4(1 - y^2))} \, dy$

Remove brackets and collect like terms

$V = 2\pi\int\limits^1_0 {4 - 4 + 8\sqrt{(1 - y^2)} +8 \sqrt{(1 - y^2)}+ 4(1 - y^2) - 4(1 - y^2)} \, dy$

$V = 2\pi\int\limits^1_0 { 16\sqrt{(1 - y^2)} \, dy$

Rewrite as:

$V = 16* 2\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy$

$V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy$

Using the calculator:

$\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy = \frac{\pi}{4}$

So:

$V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy$

$V = 32\pi * \frac{\pi}{4}$

$V =\frac{32\pi^2}{4}$

$V =8\pi^2$

Take:

$\pi = 3.142$

$V = 8* 3.142^2$

$V = 78.97731$ --- approximated

3 0

3 years ago

10 < 3x + 19<br> Help ASAP time is running out !

timurjin [86]

Answer:

-3<x

Step-by-step explanation:

10 < 3x + 19

10 - 19 < 3x

-9 < 3x

-9/3 < x

-3 < x

5 0

4 years ago

Read 2 more answers