All categories

2 years ago

How do you solve 0.3+1.4(x+1)=4.4x+8.3-3x

Mathematics

1 answer:

chubhunter [2.5K]2 years ago

7 0

Answer:

x=6.7

Step-by-step explanation:

To solve this problem is necessary to perform basic operations in order to reduce the amount of work.

This means that we need to sum coefficients of 'x' in boths side of equations, and do the same thing to free terms.

Therefore, The expression 0.3+1.4(x+1)=4.4x+8.3-3x, becomes

0.3+1.4(x+1)=4.4x+8.3-3x

1.4x+1.7=1.1x+8.3

Lets isolate x variable in one side of equality sign

1.4x+1.1x=8.3+1.7

1.5x=10

x=10/1.5

You might be interested in

Solve for y.<br> 3x + 4y = z<br> -----------------

Georgia [21]

Answer:

y= z-3x/4

Step-by-step explanation:

4y=z-3x

y=z-3x/4

3 0

3 years ago

Ms. Carson's art class made a large rectangular mural. The length of the mural is 9 feet, and it's area is 72 square feet. What

lorasvet [3.4K]

You do the reverse and divide the 2 values to find out the last one

72/9 = 8

Do you get it?

8 0

2 years ago

Read 2 more answers

Pls help!!! Which steps would you use to solve this equation? Check all that apply.

hram777 [196]

Answer:

you subtract 7 from both sides

Step-by-step explanation:

3 0

3 years ago

a grocery store display has 2 boxes of cereal on the top row,3 boxes in the next row,4 boxes in the next row, and so on. How man

sveticcg [70]

There ya go! The answer is 8 rows. Added a picture to show you my work :)

8 0

3 years ago

Read 2 more answers

Find the projection of u onto v<br> URGENT HELP

padilas [110]

Given:

Two vectors are:

$u=\left$

$v=\left$

To find:

The projection of u onto v.

Solution:

Magnitude of a vector $v=\left$ is:

$|v|=\sqrt{a^2+b^2}$

Dot product of two vector $v_1=\left$ and $v_2=\left$ is:

$v_1\cdot v_2=a_1a_2+b_1b_2$

Formula for projection of u onto v is:

$Proj_vu=\dfrac{u\cdot v}{|v|^2}v$

$Proj_vu=\dfrac{\left \cdot \left }{(\sqrt{2^2+17^2)^2}}\left$

$Proj_vu=\dfrac{0\cdot 2+6\cdot 17}{4+289}\left$

$Proj_vu=\dfrac{0+102}{293}\left$

On further simplification, we get

$Proj_vu=\dfrac{102}{293}\left$

$Proj_vu=\left$

Therefore, the projection of u onto v is $Proj_vu=\left$ .

3 0

3 years ago