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3 years ago

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Solve the system of equations

1 answer:

Nadya [2.5K]3 years ago

4 0

So we have the equations:

y - 4x = 2 --- equation 1

5y - 2x = 1 --- equation 2

(equation 2) * 2

10y - 4x = 2 --- equation 3

(equation 2) - (equation 3)

-9y = 0 so y= 0 and thus x = -1/2

Hope that helps!

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Solve the system by elimination.(show your work)

PilotLPTM [1.2K]

Answer:

x = 1 , y = 1 , z = 0

Step-by-step explanation by elimination:

Solve the following system:

{-2 x + 2 y + 3 z = 0 | (equation 1)

-2 x - y + z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x - 3 y - 2 z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y+0 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1 , y = 1 , z = 0

6 0

3 years ago

Read 2 more answers

fgiga [73]

Answer:

5

Step-by-step explanation:

sorry i dont know

4 0

3 years ago

A simple main effect analysis examines: relationships between independent and dependent variables. the overall effect of the int

Oksi-84 [34.3K]

A simple main effect analysis looks at the mean difference at each level of the independent variable.

Given With a simple main effect, the analysis examines.

Simple effects are also called the main effects of design of experiments. The main effect can be defined as the effect of the explanatory variables measured on a particular criterion or response variable of the study. It's easy to say that what happens within the level of the explanatory variables is the difference. In a factorial experiment, the simple main effect is that one independent variable is at a specified level of another. The main effect of one independent variable is averaged across the levels of the other independent variable. For a simple main effect, the results are analyzed as if a separate experiment was performed at each level of the other independent variables.

Therefore, a simple main effect analysis looks at the mean difference at each level of the independent variable.

Learn more about analysis from here brainly.com/question/15179003

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4 0

2 years ago

(4*10^2)(2*10^5) written in scientific notation

Akimi4 [234]

<span>(4*10^2)(2*10^5)
= 8 x 10^7

hope it helps</span>

8 0

3 years ago

Read 2 more answers

(sinA-cosA+1)/(sinA+cosA-1)=2(1+cosecA)

borishaifa [10]

Answer:

The trigonometrical expression is sin² A + sin A - 2 cos A - 2 cos A × sin A = 0

Step-by-step explanation:

Given Trigonometrical function as :

$\frac{sin A - cos A + 1}{sin A + cos A - 1}$ = 2 (1 + cosec A)

Or, $\frac{sin A + ( 1 - cos A)}{sin A - (1 - cos A)}$ = 2 (1 + cosec A)

,<u> Now, rationalizing </u>

$\frac{(sin A + ( 1 - cos A)) \times (sin A + (1 - cosA))}{(sin A - (1 - cos A))\times (sin A + (1 - cos A))}$ = 2 (1 + cosec A)

Or, $\frac{(sin A + (1 - cos A))^{2}}{sin^{2} - (1-cos A)^{2}}$ = 2 ( 1 + $\dfrac{1}{\textrm sinA}$

Or, $\frac{sin^{2}A + (1 - cosA)^{2} + 2 \times sin A \times (1 - cos A)}{sin^{2}A - (1 + cos^{2}A - 2 cos A)}$ = 2 ( $\dfrac{1 + sin A}{sin A}$

Or, $\frac{sin^{2}A + 1 + cos^{2}A - 2 cos A + 2 sin A - 2 sin A cos A}{sin^{2}A - 1 - cos^{2}A +2 cos A}$ = 2 ( $\dfrac{1 + sin A}{sin A}$

Or, $\frac{sin^{2}A + 1 + cos^{2}A - 2 cos A + 2 sin A - 2 sin A cos A}{sin^{2}A - (sin^{2}A + cos^{2}A) - cos^{2}A +2 cos A}$ = 2 ( $\dfrac{1 + sin A}{sin A}$

Or, $\frac{2- 2 cos A + 2 sin A - 2 sin A cos A}{- 2cos^{2}A +2 cos A}$ = 2 ( $\dfrac{1 + sin A}{sin A}$

Or, $\frac{1- cos A + sin A - sin A cos A}{- cos^{2}A + cos A}$ = 2 ( $\dfrac{1 + sin A}{sin A}$

Or, $\frac{(1- cos A) + sin A (1-cos A)}{cos A(1 - cos A)}$ = 2 ( $\dfrac{1 + sin A}{sin A}$

Or, $\frac{(1- cos A) (1 + sinA)}{cos A(1 - cos A)}$ = 2 ( $\dfrac{1 + sin A}{sin A}$

Or, $\frac{(1 + sinA)}{(cos A)}$ = 2 ( $\dfrac{1 + sin A}{sin A}$

Or, sin A + sin² A = 2 cos A (1 + sin A)

Or, sin A + sin² A = 2 cos A + 2 cos A × sin A

Or, sin² A + sin A - 2 cos A - 2 cos A × sin A = 0

So,The trigonometrical expression is sin² A + sin A - 2 cos A - 2 cos A × sin A = 0 Answer

6 0

3 years ago