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3 years ago

Can you please help me with this problem?

Mathematics

1 answer:

Hoochie [10]3 years ago

6 0

Here you have to find which each variable is, for this you start of picking one equation,

x + 2y + 6z = 4
-3x + 2y - 2 = -4
4x + 2z = 16

depending the equation you pick you multiply that by a certain number that will give you the opposite of one of the other equations,
-1(x + 2y + 6z = 4)
= -x -2y - 6z = -4

With this you add or subtract it with the equation that has the same number or variable, or both,
In this case it will be the equation,
-3x + 2y + 6z = 4
You can use this one or the third equation since both have a positive 2y which will cancel with -2y from the new equation,
-x - 2y - 6z = -4
-3x + 2y -z = -4

= -4x -7z = -8
Now you since you just eliminated the variable (y) you now have 2 variables, and the last equation has only 2 variables, meaning now you find the answer to those to equations,

-4x -7z = -8
4x + 2z = 16

= -5z = 8
Now leave the variable by itself,
z = 8/5
Now you found the variable (z), with this just substitute on one of the equations we used to find (z) so you can find (x), after that substitute those answered to on of the original equations so you can find (y)

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Step-by-step explanation:

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4 years ago

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Answer:

Step-by-step explanation:

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3 years ago

Find the additive inverse of-9x+3

miv72 [106K]

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Step-by= explanation

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3 years ago

Read 2 more answers

I got this wrong. not sure where I went wrong

oksano4ka [1.4K]

Answer:

$\left( x-2\right)^{2} +\left( y-6\right)^{2} =12$

Step-by-step explanation:

Formula ………………………………………………………………………

The equation of a circle with center (a , b) and radius r is :

$\left( x-a\right)^{2} +\left( y-b\right)^{2} =r^{2}$

===================

Now ,let’s apply what we have learnt:

The equation of the circle with center (a , b) = (2 , 6) and radius r = 2√3 is :

$\left( x-2\right)^{2} +\left( y-6\right)^{2} =\left( 2\sqrt{3} \right)^{2}$

$\Longleftrightarrow \left( x-2\right)^{2} +\left( y-6\right)^{2} =12$

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2 years ago

How do I do this question folks?

horsena [70]

Answer:

Correct answer: B

Step-by-step explanation:

Syntax: in piecewise functions such as the one attached, the "if:" section shows the domain, or x-axis values which that function pertains to.

In the graph, you can see that the graph is defined for $-3\leq x < 1$ (not-including 1 because there is an open hole there, indicating it is not part of the domain), and $2\leq x\leq 4$ .

Now that we know the domain, we can attach it to the graphs that lie on those domains.

We see that the leftmost line appears to have a positive slope and a negative y-intercept, and that the second line should have a positive y-intercept and a negative slope.

At this point, you can just start crossing off answers that don't meet this criteria.

Cheers!!

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3 years ago