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Who's good at math?
Write an equation for the line that is parallel to the given line and that passes through the given point. y=-5x+3 ; (-6,3)
Find a solution to the following system of equations. -5x+y=-5 and -4x+2y=2
Same with this one. y=3x+7 and y=x-9
Find the area of the unshaded region. (answer in simlified form) Big square is x+5 and little square inside the big one is x. x is the shaded square.
I just need to know how to do each one of them, thank you.
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I can't really finish the problems for you because that's academic dishonesty, but here's how to do each one:
1) The general form of the equation of a line is:
y = mx + b,
where m is the slope (rise/run) and b is the point at which x = 0.
This means that a line that is parallel to y = -5x + 3 will have a slope of m = -5.
You can use the point the line passes through and the slope to find the y-intercept.
y = mx + b
3 = -5(-6) + b
b = 3 - 30
b = -27
Then you put it all together in y = mx + b form.
2) To find a solution to a set of equations, there are a few ways to go. This is one way:
We need to isolate y for both equations.
-5x + y = -5 can be rearranged to show y = 5x -5
and -4x + 2y = 2 can be rearranged as 2y = 4x + 2, then divided by a common factor of 2 to show y = 2x + 1.
Now we have two equations with y on one side. Since y = y, that means that the other halves of the equations have to be equal.
This means that
5x - 5 = 2x + 1
5x - 2x = 1 + 5
3x = 6
x = 2
Now you sub x = 2 into one of your equations for y. This gives you the solution, which you will need to put in (x,y) form. You can do the other one the same way.
3) I'm assuming that the squares have a side length of x + 5 and x?
You know that the area of a square can be represented with length x length. The area of the unshaded area will equal the area of the large square minus the area of the small square.
A(Large)
= (x + 5)(x + 5)
= x^2 + 10x + 25
A(Small)
= (x)(x)
= x^2
Now subtract A(Small) from A(Large).
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