All categories

3 years ago

Linear Functions! Step by step Solution for 4(x+2)-1/2x=22 equation Thankss❤️

Mathematics

2 answers:

Svetach [21]3 years ago

7 0

4(x+2)-1/2x=22 we need to get it into the form of a+bx
4x+8-1/2x=22
3.5x=14
f(x)=-14+3.5x

Hope that helps

Elena L [17]3 years ago

6 0

4(x + 2) - 1/2x = 22
4x + 8 - 1/2x = 22
3.5x + 8 = 22
-8 -8
3.5x = 14
/ 3.5 /3.5
x = 4

You might be interested in

How many line segments would you use to draw 14 pentagons

AfilCa [17]

The ratio for this problem is 1:5.

If you are drawing 14 pentagons, you would do 14 x 5.

14 (pentagons) x 5 (line segments) = 70 line segments in total.

The answer is C) 70

6 0

2 years ago

What is the slope intercept ?

Olegator [25]

Answer:

y=mx+b

Step-by-step explanation:

that is the equation

4 0

2 years ago

You invest $350 in an account with an interest rate of 1.2% compounded continuously. How much money would be in the account afte

In-s [12.5K]

You will get about 1300 a year

6 0

3 years ago

I will give BRAINLIEST to the correct answer Find the measure of angle 6.

True [87]

Answer:

m∠6 = 116°

Step-by-step explanation:

first find x:

7x - 17 = 2x + 78

subtract 2x from both sides: 5x - 17 = 78

add 17 to both sides: 5x = 95

divide by 5: x = 19

plug x into 7x - 17

7(19) - 17 = 116

∠6 and 7x - 17 are vertical angles and therefore congruent, so m∠6 also = 116°

3 0

2 years ago

Find the limit, if it exists, or type dne if it does not exist.

Phantasy [73]

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{\left(x+23y)^2}{x^2+529y^2}$

Suppose we choose a path along the $x$ -axis, so that $y=0$ :

$\displaystyle\lim_{x\to0}\frac{x^2}{x^2}=\lim_{x\to0}1=1$

On the other hand, let's consider an arbitrary line through the origin, $y=kx$ :

$\displaystyle\lim_{x\to0}\frac{(x+23kx)^2}{x^2+529(kx)^2}=\lim_{x\to0}\frac{(23k+1)^2x^2}{(529k^2+1)x^2}=\lim_{x\to0}\frac{(23k+1)^2}{529k^2+1}=\dfrac{(23k+1)^2}{529k^2+1}$

The value of the limit then depends on $k$ , which means the limit is not the same across all possible paths toward the origin, and so the limit does not exist.

8 0

3 years ago