All categories

3 years ago

2y-3x=4 solving for y

2 answers:

6 0

<span>2y-3x=4
Add 3x to both sides
2y=3x+4
Divide both sides by 2 so that the only thing remaining on the left side is the variable y.
Final Answer: 3/2x+2</span>

d1i1m1o1n [39]3 years ago

5 0

2y-3x=4
2y=3x+4 (Added 3x to both sides)
y=3/2x+2 (Divided both sides by 2)

You might be interested in

By installing efficient water fixtures the average American can reduce water use to about 45 gallons of water per day.Using such

zimovet [89]

Hello there.

<span>By installing efficient water fixtures the average American can reduce water use to about 45 gallons of water per day.Using such water fixtures about how many gallons of water would the average American use in December
</span>
D about 2000 gallons

6 0

3 years ago

Find all solutions of the given system of equations and check your answer graphically. HINT [See Examples 1-4.] (If there is no

Likurg_2 [28]

Answer:

$Infinitely\ many\ solutions\ exist.\\\\Solutions\ are\ (x,\frac{3}{4}x-1)$

Step-by-step explanation:

$Given\ equations\ are\\\\3x-4y=4.................eq(1)\\\\9x-12y=12..............eq(2)\\\\divide\ eq(2)\ by\ 3\\\\\frac{1}{3}(9x-12y=12)\\\\\Rightarrow 3x-4y=4\\\\Hence\ equations\ represent\ the\ same\ line.\\Hence\ Infinitely\ many\ solutions\ exist.\\\\3x-4y=4\\\\4y=3x-4\\\\y=\frac{3}{4}x-1\\\\Solutions\ are\ (x,\frac{3}{4}x-1)$

3 0

3 years ago

Find the volume of the cubes or rectangular prisms in cubic centimeters.

DanielleElmas [232]

a) Volume of Rectangular prism is $\frac{36}{125}\,\,cm^3$

b) Volume of cube is $\frac{8}{125}\,\,cm^3$

c) Volume of cube is $\frac{1}{125}\,\,cm^3$

d) Volume of Rectangular prism is $\frac{6}{125}\,\,cm^3$

Step-by-step explanation:

Part a)

Volume of rectangular prism with

length= $\frac{3}{5}\,\,cm$

width= $\frac{4}{5}\,\,cm$

height = $\frac{3}{5}\,\,cm$

The formula used to find Volume of rectangular prism is:

$Volume\,\,of\,\,rectangular\,\,prism=length*width*height$

Putting values:

$Volume\,\,of\,\,rectangular\,\,prism=\frac{3}{5} *\frac{4}{5} *\frac{3}{5}\\Volume\,\,of\,\,rectangular\,\,prism=\frac{36}{125}\,\,cm^3$

So, Volume of Rectangular prism is $\frac{36}{125}\,\,cm^3$

Part b)

Volume of cube with side length of $\frac{2}{5}\,\,cm$

The formula used to find Volume of cube is:

$Volume\,\,of\,\,cube=length^3$

Putting value of length and finding volume:

$Volume\,\,of\,\,cube=length^3\\Volume\,\,of\,\,cube=(\frac{2}{5})^3\\ Volume\,\,of\,\,cube=\frac{8}{125}\,\,cm^3$

So, Volume of cube is $\frac{8}{125}\,\,cm^3$

Part c)

Volume of cube with side length of $\frac{1}{5}\,\,cm$

The formula used to find Volume of cube is:

$Volume\,\,of\,\,cube=length^3$

Putting value of length and finding volume:

$Volume\,\,of\,\,cube=length^3\\Volume\,\,of\,\,cube=(\frac{1}{5})^3\\ Volume\,\,of\,\,cube=\frac{1}{125}\,\,cm^3$

So, Volume of cube is $\frac{1}{125}\,\,cm^3$

Part d)

Volume of rectangular prism with

length= $\frac{1}{5}\,\,cm$

width= $\frac{2}{5}\,\,cm$

height = $\frac{3}{5}\,\,cm$

The formula used to find Volume of rectangular prism is:

$Volume\,\,of\,\,rectangular\,\,prism=length*width*height$

Putting values:

$Volume\,\,of\,\,rectangular\,\,prism=\frac{1}{5} *\frac{2}{5} *\frac{3}{5}\\Volume\,\,of\,\,rectangular\,\,prism=\frac{6}{125}\,\,cm^3$

So, Volume of Rectangular prism is $\frac{6}{125}\,\,cm^3$

Keywords: Volume

Learn more about Volume at:

#learnwithBrainly

6 0

3 years ago

Identify the number as real, complex, pure imaginary, or nonreal complex. (More than one of these descriptions may apply.) -9i

Agata [3.3K]

Answer:

Step-by-step explanation:

Me no hablas Inglish sorry pal

8 0

3 years ago

Find the equation of the plane that is parallel to the vectors left angle 3 comma 0 comma 3 right angle and left angle 0 comma 1

Sholpan [36]

Answer:

$x + 3y -z - 3 = 0$

Step-by-step explanation:

We have to find the equation of plane that is parallel to the vectors

$\langle 3,0,3\rangle, \langle0,1,3\rangle$

The plane also passes through the point (2,0,-1).

Hence, the equation of plane s given by:

$\displaystyle\left[\begin{array}{ccc}x-2&y-0&z+1\\3&0&3\\0&1&3\end{array}\right]\\\\=(x-2)(0-3) - (y-0)(9-0) + (z+1)(3-0)\\=-3(x-2)-9y+3(z+1)\\\Rightarrow -3x + 6 - 9y + 3z + 3 = 0\\\Rightarrow 3x + 9y -3z -9 = 0\\\Rightarrow x + 3y -z - 3 = 0$

It is the required equation of plane.

4 0

3 years ago