In ( -5 , 4) ,x = -5 and y = 4So , according to above figure ( -5 , 4) will be in second quadrant .
Math question Really important (200 points)?
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(0,3) is a point on the [math]y[/math] axis and therefore not considered to be in any of the four quadrants. The same would apply if the point is on the [math]x[/math] axis.In general, the quadrants are given byFirst: [math](x,y)=(+,+)[/math]Second: [math](x,y)=(-,+)[/math]Third: [math](x,y)=(-,-)[/math]Fourth: [math](x,y)=(+,-)[/math]where [math]+[/math] denotes a positive real number and [math]-[/math] denotes a negative real number
Hello,The point (3,-2) is lie in the Fourth Quadrant.Illustration is as follows :In the First Quadrant, both x and y are positive. Following figure will show you in details:2. In Second Quadrant x axis pooints are negative and y axis points are positive.3. In third Quadrant x axis points are negativeand y axis points are also negative.4. In Fourth Quadrant x axis points are positive and y axis points are negative so thePoint (3, -2) lie in the quadrant 4.See in following picture:Thank You…!!☺
Math SAT question prep help?
The problem discusses "the product of the x coordinate and the y coordinate." Let me give you some examples. If the point is (1,3), then the product is 1*3 = 3. If the point is (-2,4), then the product is -2*4 = -8. If the point is (0,3), then the product is 0*3 = 0. You want the points on your line to have this product less than or equal to 0. That is, if we describe a point on the line by (x,y), then we need x*y <= 0. You should remember from pre-algebra that this occurs if x<=0 or y<=0, but not if x<0 AND y<0. So some examples of points that would work: (-1, 1) (-1, 0) (0, 0) (3, -4) (43208762, 0) And some points that would not work: (93, 2) (-3, -4) etc. Geometric interpretation: If you plot some of these, or just have some good intuition, you will see that all the points that work are in the second and fourth quadrants of the graph. So you want a line which passes through these quadrants, which would be any line with a negative slope and 0 intercept. (For instance, y=-2x, or y=-238947x, or y=-x). Algebraic interpretation: You want the y-coordinate and the x-coordinate to be inversely related. That is, if one is positive, then the other is negative. So it makes sense that it equation relating them would have a minus sign. Can you see from the equation y=-2x that if x is positive, y is negative, and vice versa?
What quadrant is the point (-3, 4) located in?
I'm not quite sure what you're asking in the first two lines. Are you just supposed to multiply them out? If so, it's pretty simple: 5(2x - 7) = 10x - 35 (4a - 3)6 = 24a - 18 What quadrant is the point (-3, 4) located in? Quadrant 2. -3(5n - 4) = -15n + 12 Evaluate the following expression for n = -2 and q = -8 4n – 6q + 3(2n – 4q) + 5n 4(-2) - 6(-8) + 3(2(-2) - 4(-8)) + 5(-2), just replace all the n's with -2 and all the q's with -8 -8 + 14 + 3(-4 + 32) - 10 6 + 3(28) -10 6 + 84 - 10 80 Solve: 4r – 12 = -16 4r = -4 (add 12 to each side) r = -1 (divide both sides by 4) Sorry if I didn't understand the 1st, 2nd, and 4th question. Hope this helps.
GO TO THE ANSWERS. Everyone is saying eliminate the most as you could with as little work as you can, and, while that is correct, is impossible without at least giving it some thought.My advice for the SAT as far as speed is concerned is to take the most systematic approach as possible. Experience says do the operations with 0, but you can also start with positive numbers because operations on negative numbers often do not correlate well with high school student's intuitions. 0->3. Gotta be +3 ad the end. Now we're left with A and B, a linear expression and one of degree 3. Where will that difference most likely show up, in larger numbers. Jump to 4->15.4^3 is simply too large to be 15. The answer is B
Ordinates, the y-coordinates of a point, on the left side of the y axis in the II & III quadrants are all negative, while the abscissas are negative below the x-axis in the III & IV quadrants. Thus, in the I quadrant both coordinates are positive, in the III quadrant both are negative, & in the II & IV quadrants they are of opposite sign. On the y-axis the abscissa is always zero while on the x-axis the ordinate is always zero.
The point (3,-2) is in a quadrant where tan x and csc x are both negative.?
Yes, tan x and csc x are both negative in the 4th quadrant. | | --------|--------- | . <- | The point being there, tan x becomes -2 / 3 and csc x becomes hypotenuse/-2 . The hypotenuse being +ve, tan x and csc x are -ve.