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3 years ago

Good integer examples that would equal 36

Mathematics

1 answer:

alex41 [277]3 years ago

4 0

-36 . negative 36 is an example because it has the same value as 36 because they are both the same distance away from 0

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X-3=10 does anyone know this

vovikov84 [41]

Answer:

Are you trying to solve for x?

x= 13

just kick the 3 to the other side, so a negative 3 would become a positive 3:

x= 10 + 3

x= 13

You can also use cymath it gives you the answer and explanation.

Hope this helps.

5 0

3 years ago

What is the rate of change? the second 1 and second 4?

Mumz [18]

Idek but I do know that 12x8= 72

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2 years ago

If 5 gallons cost $ 6.60 how much do 29 gallons of gasoline cost

ikadub [295]

Answer:

Should be 38.28

Step-by-step explanation:

Dividing 6.60 by 5 is $1.32, aka the cost of one gallon. multiply by 29, answer is that it costs $38.28 to get 29 gallons of gasoline.

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2 years ago

<img src="https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D1%7D%20%5Catop%20%7Bx-2y%3D4%7D%7D%20%5Cright.%20%5C%5C%5Clef

brilliants [131]

Answer:

(a) x=2, y=-1

(b) x=2, y=2

(c) $\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) x=-2, y=-7

Step-by-step explanation:

Cramer's Rule

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

We call the determinant of the system

$\Delta=\begin{vmatrix}a &b \\c &d \end{vmatrix}$

We also define:

$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

$\displaystyle x=\frac{\Delta_x}{\Delta}$

$\displaystyle y=\frac{\Delta_y}{\Delta}$

(a) The system to solve is

$\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.$

Calculating:

$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

$\Delta_y=\begin{vmatrix}1 &1 \\1 &4 \end{vmatrix}=4-3=3$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1$

The solution is x=2, y=-1

(b) The system to solve is

$\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.$

Calculating:

$\Delta=\begin{vmatrix}4 &-1 \\1 &-1 \end{vmatrix}=-4+1=-3$

$\Delta_x=\begin{vmatrix}6 &-1 \\0 &-1 \end{vmatrix}=-6-0=-6$

$\Delta_y=\begin{vmatrix}4 &6 \\1 &0 \end{vmatrix}=0-6=-6$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2$

The solution is x=2, y=2

$\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.$

Calculating:

$\Delta=\begin{vmatrix}-1 &2 \\1 &2 \end{vmatrix}=-2-2=-4$

$\Delta_x=\begin{vmatrix}0 &2 \\5 &2 \end{vmatrix}=0-10=-10$

$\Delta_y=\begin{vmatrix}-1 &0 \\1 &5 \end{vmatrix}=-5-0=-5$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}$

The solution is

$\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) The system to solve is

$\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.$

Calculating:

$\Delta=\begin{vmatrix}6 &-1 \\4 &-2 \end{vmatrix}=-12+4=-8$

$\Delta_x=\begin{vmatrix}-5 &-1 \\6 &-2 \end{vmatrix}=10+6=16$

$\Delta_y=\begin{vmatrix}6 &-5 \\4 &6 \end{vmatrix}=36+20=56$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7$

The solution is x=-2, y=-7

4 0

3 years ago

What is the equation of the line that passes through the point (6,14) and is parallel to the line with the following equation? y

Gekata [30.6K]

Answer:

$y=\displaystyle-\frac{4}{3}x+22$

Step-by-step explanation:

Hi there!

What we need to know:

Linear equations are typically organized in slope-intercept form: $y=mx+b$ where m is the slope and b is the y-intercept
Parallel lines always have the same slope (m)

Determine the slope (m):

$y=\displaystyle-\frac{4}{3}x -1$

The slope of the given line is $\displaystyle-\frac{4}{3}$ , since it is in the place of m in y=mx+b. Because parallel lines always have the same slope, the slope of a parallel line would also be $\displaystyle-\frac{4}{3}$ . Plug this into y=mx+b:

$y=\displaystyle-\frac{4}{3}x+b$

Determine the y-intercept (b):

$y=\displaystyle-\frac{4}{3}x+b$

To find the y-intercept, plug in the given point (6,14) and solve for b:

$14=\displaystyle-\frac{4}{3}(6)+b\\\\14=-8+b\\b=22$

Therefore, the y-intercept of the line is 22. Plug this back into $y=\displaystyle-\frac{4}{3}x+b$ :

$y=\displaystyle-\frac{4}{3}x+22$

I hope this helps!

5 0

3 years ago