Answer:
- see the attachment
- (x, y) = (1, 1)
Step-by-step explanation:
1. Since you have y > ..., the boundary line is dashed and the shading is above it (for y-values greater than the values on the line). The boundary line is ...
y = 2x+3
which has a y-intercept of 3 and a slope (rise/run) of 2. A graph is attached.
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2. You can add the two equations to eliminate y:
(3x +y) +(x -y) = (4) +(0)
4x = 4
x = 1
1 - y = 0 . . . . substitute into the the second equation
1 = y . . . . . . . add y
The solution is (x, y) = (1, 1).
If they meant 10 sets of push-ups, then she did 45. if they just meant 10 push-ups, it is 25.
Rule: second differences are 2.
Selection C is appropriate.
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First differences are the sequence of differences of the given numbers:
.. 3 -8 = -5
.. 0 -3 = -3
.. -1 -0 = -1
.. 0 -(-1) = 1
Second differences are the sequence of differences of these:
.. -3 -(-5) = 2
.. -1 -(-3) = 2
.. 1 -(-1) = 2
All of these are 2, so the sequence can be described by a *second* order polynomial (because *second* differences are a non-zero constant).
D. -4
sq rt of 3x+28-8=x
add 8 to both sides
sq rt of 3x+28 = x+8
square both sides of the equation
3x + 28 = x^2 + 16x + 64
subtract x^2+16x+64 from both sides
group common terms
-13x-36-x^2=0
rearrange and put in quadratic form
x= -13 +/- sq rt of 13^2-4(1)(36) over 2(1)
subtract 169 by 144
x= -12 +/- sq rt of 25 over 2
solve for both sides
x = -13+5 over 2
=-4
x= -13-5 over 2
= -9
now plug in x
sq rt of 3(-4)+28 -8=-4
sq rt of 3(-9)+28 -8= -9
-4 = -4
but
-7 does not = -9
so -4 is the solution
