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3 years ago

The area of a rectangle is 20 cm

Mathematics

1 answer:

Nezavi [6.7K]3 years ago

4 0

Answer:

W=2cm

Step-by-step explanation:

The area of a rectangle is 20 cm, with the length being 8 more than the width, two common factors to 20 using this criteria would be 10 and 2. 10 As the length as it is 8 more than the 2 as the width.

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I will MARK THE CROWN IF SOMEONE ANSWERS THESES

kirza4 [7]

Answer:

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the third picture answer is

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8 0

3 years ago

A cylinder has a base diameter of 20 meters and a height of 10 meters. What is it’s volume in cubic meters, to the nearest tenth

Korvikt [17]

3141.6 cubic meters

Step-by-step explanation:

The volume of a cylinder is given by;

πr²h

π * (²⁰/₂)² * 10

π * 100 * 10

= 3141.6

4 0

4 years ago

Convert from Decimal Notation to Scientific Notation

Bingel [31]

Answer:

Hence, 0.0000087 can be expressed as $8.71 \times 10^{-6}$ .

Step-by-step explanation:

- Given 0.00000871

- Move the decimal point so that the factor is greater than or equal to 1 but less than 10 .

- Then count n and write no. as product with a power of 10 .

Step 1 of 1

Consider 0.00000871,

8.71

Decimal has moved 6 places to right.

The power will be negative as the original no. is less than 1 .

$\begin{array}{r}8.71 \times 10^{-6} \\0.0000087=8.71 \times 10^{-6}\end{array}$

3 0

2 years ago

What is 70g increased by 10%

rosijanka [135]

Answer:

77g

Step-by-step explanation:

10 % of 70 = 7

70 + 7 = 77

70 / 10 = 7 ( 10%)

8 0

3 years ago

Read 2 more answers

Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable

bekas [8.4K]

Answer:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

Step-by-step explanation:

First step: I'm going to solve our substitution for y:

$u=\frac{y}{x}$

Multiply both sides by x:

$ux=y$

Second step: Differentiate the substitution:

$u'x+u=y'$

Third step: Plug in first and second step into the given equation dy/dx=f(x,y):

$u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}$

$u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}$

We are going to simplify what we can.

Every term in the fraction on the right hand side of equation contains a factor of $x^2$ so I'm going to divide top and bottom by $x^2$ :

$u'x+u=\frac{u+u^2}{3+u^2}$

Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:

Subtract u on both sides:

$u'x=\frac{u+u^2}{3+u^2}-u$

Find a common denominator: Multiply second term on right hand side by $\frac{3+u^2}{3+u^2}$ :

$u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}$

Combine fractions while also distributing u to terms in ( ):

$u'x=\frac{u+u^2-3u-u^3}{3+u^2}$

$u'x=\frac{-u^3+u^2-2u}{3+u^2}$

Third step: I'm going to separate the variables:

Multiply both sides by the reciprocal of the right hand side fraction.

$u' \frac{3+u^2}{-u^3+u^2-2u}x=1$

Divide both sides by x:

$\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}$

Reorder the top a little of left hand side using the commutative property for addition:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

The expression on left hand side almost matches your expression but not quite so something seems a little off.

5 0

3 years ago