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3 years ago

The sum of two numbers is 20,one number is 4 less than the other.find the number

Mathematics

2 answers:

kogti [31]3 years ago

5 0

I’ve attached my work
Hope this helps!

Marysya12 [62]3 years ago

4 0

X-4+x=20

2x=20+4
2x=24
X=12
Number is 12 because 12-4 is 8 and 8+12 is 20

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If 3% of a number equals 1, find 12% of that number.

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Answer:

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Step-by-step explanation:

.03n = 1

n = 1/.03

.12 x (1/.03) = 4

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13) Find the value for x and y. Someone help me please thanks

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Answer:

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Step-by-step explanation:

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Solve the following system of equations. Enter the y-coordinate of the solution. Round your answer to the nearest tenth.

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3 years ago

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What is the measure of LXZW? 850 150° Y 65° 95 Z 215° W 30°

Semenov [28]

Answer:

so the answer is 95°

Step-by-step explanation:

The two angles on a line add up to 180° or are supplementary, so we can find the inside angle x by doing 180-150=30°. Now that we know inside x is 30, if we add 30+65, we get 95, and 180-95=85°. Then to find angle xzw, just do 180-85 to get 95°.

3 0

3 years ago

Solve using elimination x+y-2z=8 5x-3y+z=-6 -2x-y+4z=-13

Free_Kalibri [48]

So here is your answer with LaTeX issued format interpretation. Full process elucidated briefly, below:

$\begin{alignedat}{3}x + y - 2z = 8 \\ 5x - 3y + 2 = - 6 \\ - 2x - y + 4z = - 13 \end{alignedat}$

For this equation to get obtained under the impression of those variables we have to eliminate them individually for moving further and simplifying the linear equation with three variables along the axis.

Multiply the equation of x + y - 2z = 8 by a number with a value of 5; Here this becomes; 5x + 5y - 10z = 40; So:

$\begin{alignedat}{3}5x + 5y - 10z = 40 \\ 5x - 3y + z = - 6 \\ - 2x - y + 4z = - 13 \end{alignedat}$

Pair up the equations in a way to eliminate the provided variable on our side, that is; "x":

5x - 3y + z = - 6

-

5x + 5y - 10z = 40
______________

- 8y + 11z = - 46

Therefore, we are getting.

$\begin{alignedat}{3}5x + 5y - 10z = 40 \\ - 8y + 11z = - 46 \\ - 2x - y + 4z = - 13 \end{alignedat}$

Multiply the equation of 5x + 5y - 10z = - 40 by a number with a value of 2; Here this becomes; 10x + 10y - 20z = 80.

Multiply the equation of - 2x - y + 4z = - 13 by a number with a value of 5; Here this becomes; - 10x - 5y + 20z = - 65; So:

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 8y + 11z = - 46 \\ - 10x - 5y + 20z = - 65 \end{alignedat}$

Pair up the equations in a way to eliminate the provided variables on our side, that is; "x" and "z":

- 10x - 5y + 20z = - 65

+
10x + 10y - 20z = 80
__________________

5y = 15

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 8y + 11z = - 46 \\ 5y = 15 \end{alignedat}$

Multiply the equation of - 8y + 11z = - 46 by a number with a value of 5; Here this becomes; - 40y + 55z = - 230.

Multiply the equation of 5y = 15 by a number with a value of 8; Here this becomes; 40y = 120; So:

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 40y + 55z = - 690 \\ 40y = 120 \end{alignedat}$

Pair up the equations in a way to eliminate the provided variables on our side, that is; "y":

40y = 120

+

- 40y + 55z = - 230
_________________

55z = - 110

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 40y + 55z = - 230 \\ 55z = - 110 \end{alignedat}$

Solving for the variable of 'z':

$\mathsf{55z = - 110}$

$\bf{\dfrac{55z}{55} = \dfrac{-110}{55}}$

Cancel out the common factor acquired on the numerator and denominator, that is, "55":

$z = - \dfrac{\overbrace{\sout{110}}^{2}}{\underbrace{\sout{55}}_{1}}$

$\boxed{\mathbf{z = - 2}}$

Solving for variable "y":

$\mathbf{\therefore \quad - 40y - 55 \big(- 2 \big) = - 230}$

$\mathbf{- 40y - 55 \times 2 = - 230}$

$\mathbf{- 40y - 110 = - 230}$

$\mathbf{- 40y - 110 + 110 = - 230 + 110}$

Adding the numbered value as 110 into this equation (in previous step).

$\mathbf{- 40y = - 120}$

Divide by - 40.

$\mathbf{\dfrac{- 40y}{- 40} = \dfrac{- 120}{- 40}}$

$\mathbf{y = \dfrac{- 120}{- 40}}$

$\boxed{\mathbf{y = 3}}$

Solve for variable "x":

$\mathbf{10x + 10y - 20z = 80}$

$\mathbf{Since, \: z = - 2; \quad y = 3}$

$\mathbf{10x + 10 \times 3 - 20 \times (- 2) = 80}$

$\mathbf{10x + 10 \times 3 + 20 \times 2 = 80}$

$\mathbf{10x + 30 + 20 \times 2 = 80}$

$\mathbf{10x + 30 + 40 = 80}$

$\mathbf{10x + 70 = 80}$

$\mathbf{10x + 70 - 70 = 80 - 70}$

$\mathbf{10x = 10}$

Divide by this numbered value $\mathbf{10}$ to get the final value for the variable "x".

$\mathbf{\dfrac{10x}{10} = \dfrac{10}{10}}$

The numbered values in the numerator and the denominator are the same, on both the sides. This will mean the "x" variable will be left on the left hand side and numbered values "10" will give a product of "1" after the division is done. On the right hand side the numbered values get divided to obtain the final solution for final system of equation for variable "x" as "1".

$\boxed{\mathbf{x = 1}}$

Final solutions for the respective variables in the form of " (x, y, z) " is:

$\boxed{\mathbf{\underline{\Bigg(1, \: \: 3, \: \: - 2 \Bigg)}}}$

Hope it helps.

8 0

3 years ago

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