1. solve for x in first to just find nterms of y
x=12-y
sub that for x in other one
first divide that one by 2 to make it easier
2x-y=18
2(12-y)-y=18
24-2y-y=18
24-3y=18
minus 24 both sides
-3y=-6
divide -3
y=2
D is answer
2.
find intersection of x+y=5 and -2x+3y=6
or easiers, just test the intersections
wait, they all intersect in the same place
this is super easy
pick ANY point that is in that reigon and see if it is true
A, if we pick lets say (5,5)
that is false for x+y<5
not A
B
pick (10,0)
false for first one
not B
C
pick (0,0)
true for first
false for 2nd
not C
D. pick (-10,0)
true for first
true for 2nd
answer is D
answer is D for both questions
(c+3)-2c-(1-3c)=2
c+3-2c-1+3c=2
2c+2=2
2c=0
c=0
<span>In this case, the value of the first 3 (the ten-thousands) has a value of 30,000. The 3 next to it, in the hundred-thousands place, has a value of 300,000. To compare the two, the 3 on the right has a value one-tenth as much as that on the left.</span>
So for this question, we're asked to find the quadrant in which the angle of data lies and were given to conditions were given. Sign of data is less than zero, and we're given that tangent of data is also less than zero. Now I have an acronym to remember which Trig functions air positive in each quadrant. . And in the first quadrant we have that all the trig functions are positive. In the second quadrant, we have that sign and co seeking are positive. And the third quadrant we have tangent and co tangent are positive. And in the final quadrant, Fourth Quadrant we have co sign and seeking are positive. So our first condition says the sign of data is less than zero. Of course, that means it's negative, so it cannot be quadrant one or quadrant two. It can't be those because here in Quadrant one, we have that all the trick functions air positive and the second quadrant we have that sign. If data is a positive, so we're between Squadron three and quadrant four now. The second condition says the tangent of data is also less than zero now in Quadrant three. We have that tangent of data is positive, so it cannot be quadrant three, so our r final answer is quadrant four, where co sign and seek in are positive.
2х3+10х+2у2-х-у=2х3+9х+2у2-у=2*3in3+9*3+2*5in2-5=2*27+27+2*25-5=54+27+50-5=126
