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LUCKY_DIMON [66]
3 years ago
12

Solve the system of equations x+2y=-16 and x+6y=-28 by combining the equations.

Mathematics
2 answers:
baherus [9]3 years ago
8 0

Answer:

x = -10

y = -3

Step-by-step explanation:

When combining the equations, we must ensure that one of the two variables will cancel out. The easiest way to do this will be to multiply one of the equations by -1, such that the x will cancel out once they are added together. This can be done with either equation, but I will do it for the first one.

This gives us the system:

\left \{ {{-x-2y=16} \atop {x+6y=-28}} \right.

We can then add the two equations by combining each of the terms:

(x - x) + (6y - 2y) = (-28 + 16)\\(0) + (4y) = (-12)\\4y = -12\\y = -3

Now that we have the value for y, we substitute it back into any of the original equations and solve for x:

x + 2y = -16\\x + 2(-3) = -16\\x - 6 = -16\\x = -10

To check our work, we substitute both x and y into the other original equation:

x + 6y = -28\\(-10) + 6(-3) = -28\\-10 - 18 = -28\\-28 = -28

Gnesinka [82]3 years ago
3 0

Answer:

x= -18 y=-5/3

Step-by-step explanation:

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Triangles A C B and M Q R are shown. Sides A B and M R are congruent. Angles C A B and M R Q are 42 degrees. Angle C B A is 53 d
miss Akunina [59]

By ASA postulate it can be say that the triangle ABC and triangle MRQ are congruent

<h3>What is Angle Sum Property?</h3>

The sum of all three angles(interior) of a triangle is 180 degrees, and the exterior angle of a triangle measures the same as the sum of its two opposite interior angles.

Using Angle Sum Property, in ΔMRQ

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4 0
1 year ago
PLEASE I REALLY NEED HELP!
vagabundo [1.1K]

Answer:

\frac{3}{28}

Step-by-step explanation:

Probability (P) is calculated as

P = \frac{requiredoutcome}{count}

The first required outcome is a red sweet from a total of 3 + 5 = 8

P( red) = \frac{3}{8}

There are now 2 red left and a count of 7, since 1 has been eaten, thus

P( second red ) = \frac{2}{7}

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6 0
3 years ago
For the equation given below, evaluate y′ at the point (−2,2)<br> xe^y−4y=2x−4−2e^2
jeka94
Implicit differentiation
chain rule is important here

I'll show the steps partially
e^y+xe^yy'-4y'=2
xe^yy'-4y'=2-e^y
y'(xe^y-4)=2-e^y
y'=\dfrac{2-e^y}{xe^y-4}
now evaluate for (-2,2)
x=-2 and y=2
y'=\dfrac{2-e^2}{-2e^2-4}
y'=\dfrac{2-e^2}{-2e^2-4}
y'=\dfrac{e^2-2}{2e^2+4}
that's it, simplest form
3 0
3 years ago
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