One side of a square is 8 centimeters long what is the area

2ab - ab + 2a + 3b + 2ab<br> what is this simplified ?

shtirl [24]

Answer: The answer is 3ab + 2a + 3b.

Step-by-step explanation:

So we are given this equation:

2ab - ab + 2a + 3b + 2ab

First we have to add like terms:

2ab + 2ab - ab + 2a + 3b

4ab - ab + 2a + 3b

3ab + 2a + 3b

Here is your answer!

4 0

3 years ago

Read 2 more answers

The ratio of boys to girls in a class is 5:3. There are 32 students

Bezzdna [24]

Answer:

8more boys

Step-by-step explanation:

girls: 32*3/8= 12

boys: 32-12=20

20-12=8

7 0

3 years ago

sum of the angle measures of a triangle is 180° suppose that one angle in a triangle has a degree measure of 4X -3 and another h

Karo-lina-s [1.5K]

you would add 4x-3 to 6x-6 and equal it to 180° which is 6x+4x-6-3=180° which is 10x-9=180° then use inverse operation to solve it.

Answer: 18.9

6 0

3 years ago

Verify the identity (tan x + 1)^2 + (tan x-1)^2= 2 sec^2 x

Elina [12.6K]

$(\tan x+1)^2+(\tan x-1)^2=2\sec^2x\\\\\text{use}\ \tan x=\dfrac{\sin x}{\cos x}\\\\L_s=\left(\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\cos x}\right)^2+\left(\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\cos x}\right)^2\\\\=\left(\dfrac{\sin x+\cos x}{\cos x}\right)^2+\left(\dfrac{\sin x-\cos x}{\cos x}\right)^2\\\\=\dfrac{(\sin x+\cos x)^2}{\cos^2x}+\dfrac{(\sin x-\cos x)^2}{\cos^2x}\\\\\text{use}\ (a\pm b)^2=a^2\pm2ab+b^2$

$=\dfrac{\sin^2x+2\sin x\cos x+\cos^2}{\cos^2x}+\dfrac{\sin^2x-2\sin x\cos x+\cos^2}{\cos^2x}\\\\=\dfrac{\sin^2x+2\sin x\cos x+\cos^2+\sin^2x-2\sin x\cos x+\cos^2}{\cos^2x}\\\\=\dfrac{2\sin^2x+2\cos^2x}{\cos^2x}=\dfrac{2(\sin^2x+\cos^2x)}{\cos^2x}\\\\\text{use}\ \sin^2x+\cos^2x=1\\\\=\dfrac{2(1)}{\cos^2x}=2\cdot\dfrac{1}{\cos^2x}=2\left(\dfrac{1}{\cos x}\right)^2\\\\\text{use}\ \sec x=\dfrac{1}{\cos x}\\\\=2(\sec^2x)=2\sec^2x=R_s\\\\L_s=R_s\Rightarrow The\ identity$