All categories

3 years ago

X + 2 + x = 22 I need to show my work. Please help. No false answers please.

Mathematics

2 answers:

Minchanka [31]3 years ago

3 0

2x + 2 = 22
2x = 20
X = 10

katrin2010 [14]3 years ago

3 0

Answer:

x = 10

Step-by-step explanation:

First you need to add like terms (x + x), which would give you 2x.

2x + 2 = 22

Then subtract 2 from both sides, causing it to cancel out one side, and 22 - 2, on the other side:

2x + 2 = 22

-2 -2

22 - 2 is 20, giving you:

2x = 20

Then divide both sides by 2:

2x/2 = 20/2

And you should get x = 10 as your answer.

You might be interested in

How do I do number three help pleaseeeeeeee

posledela

I got 36. I did 5-2=3, 3•180=540, 540/5=108, 180-108=72, and 72/2= 36

4 0

3 years ago

– 6 x + 4 < 29 answer?

svlad2 [7]

Answer:

its 847>×+18 enjoy hope it work

4 0

3 years ago

24 divided by 4 times d

nadya68 [22]

Answer:

i think 6 not sure tho sorry if its worng

3 0

3 years ago

What is two thirds or 2/3 of the whole 120

Zepler [3.9K]

It would be 80 because 2 times 120 is 240. 240 divided by 3 is 80

3 0

3 years ago

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$

6 0

3 years ago