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MaRussiya [10]
3 years ago
5

If you were to solve the following system of equations by using a matrix,

Mathematics
1 answer:
Studentka2010 [4]3 years ago
4 0

Answer:

I believe it is C but it's good to double check

Step-by-step explanation:

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5/6 y = 12 <br> What’s the equation for y
mr_godi [17]

Answer:

y = \frac{72}{5}

Step-by-step explanation:

Multiply each side by 6/5

\frac{6}{5}(\frac{5}{6}y ) = 12(\frac{6}{5} )

Simplify

y = \frac{72}{5}

8 0
3 years ago
What is the complex conjugate of 3i - 2?
andreev551 [17]
Use m a t h w a y to find this answer very easy 3i - 2
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3 years ago
I have two questions to solve<br><br> -3x+10=34<br><br> 13x-4=11x+10-8
Marina CMI [18]

-3x + 10 = 34

Subtract 10 from both sides:

-3x = 24

Divide both sides by -3:

X = 24 / -3

X = -8

musiclover10045

Middle School Mathematics 39+20 pts

I have two questions to solve

-3x+10=34

13x-4=11x+10-8

Combine like terms on the right side:

13x -4 = 11x + 2

Subtract 11x from both sides:

2x -4 = 2

Add 4 to both sides:

2x = 6

Divide both sides by 2:

X = 6/2

X = 3

5 0
3 years ago
Read 2 more answers
630<br> Solve for &lt;2.<br> *2 = [?]°<br> 142<br> 57°123°<br> please explain as well
AlexFokin [52]

Step-by-step explanation:

the total angles in the triangle are 180°

so,

63° + 57° + ∠2 = 180°

120° + ∠2 = 180°

∠2 = 180° - 120°

∠2 = 60°

<h3>#CMIIW</h3>

7 0
3 years ago
At the beginning of its route, a garbage truck accelerates at 10ms2 as it is driven down the road. By the end of the route, the
Dmitry [639]

Answer:

2.5\ m/s^2

Step-by-step explanation:

Newton's Second Law: Force on a body is equal to the product of mass and acceleration of the centre of the mass of the body.

Force(F)=Mass(m)\times Acceleration(a)

Initially:

Let\ mass=m\ kg\\\\acceleration=10\ m/s^2\\\\Force(F)=m\times 10=10m

At the end of the road:

Let\ mass=4\times m\ kg\\\\acceleration=a\ m/s^2\\\\Force(F)=4m\times a=4ma

Both\ forces\ are\ equal\\\\4ma=10m\\\\divide\ both\ the\ sides\ by\ m\\\\4a=10\\\\a=\frac{10}{4}\\\\a=2.5\ m/s^2

8 0
3 years ago
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