What are the coordinates of the following point?
The first number of the pair in each is the x coordinate & the second is the y coordinate.
What are the coordinates of point P?
This is how mi teacher taught me first you look at the two equations, if one of them has a variable that has a co-efficient of one then focus on that one equation first. Isolate the variable, and make sure the variable is positive. ( also, label them to be the first equation and the second equation) EX: 4x - 3y = 6 equation 1 4x- y = 10 equation 2 therefore 4X - 10 = y then, take the answer and sub it into the other equation: EX: from equation 2, sub y = (4X - 10) in equation 1 4x - 3y = 6 4x - 3(4x-10) = 6 4x - 12x +30 = 6 4x - 12x = 6 - 30 -8x = -24 -8x/-8 = -24/-8 therefore, x = 3 Then, take that answer, and sub it into either equation, which ever one you want EX: sub x = 3 in equation 2 4x - y = 10 4(3) - y = 10 12 - y = 10 12 - 10 = y therefore, 2 = y Therefore, Point P should be at (x, y) = (3, 2) If you want to do a check, then take the equation THAT WAS NOT USED in the last solve, which would be equation 1, and put it into check form with the LS and RS ( make sure to state what you are checking and everything that you do such as the from equation...sub in equation... i lost marks on tests because i didnt) EX: Check (x,y) = (3, 2) in equation 1 Ls: 4x-3y Rs = 6 =4(3) - 3(2) =12-6 =6 Since Ls=Rs, Therefore (x,y) = (3,2) there are also symbols for since and therefore to make it easier instead of writting the actual work since, and therefore Ex: since: . . . Therefore: . . . im sorry i cant do them properly .... hope it helps ^^
What are the coordinates of the following:?
That's only one line. You need two lines to find a point of intersection.
Of the following coordinates which is a point on the graph?
in A through E the coordinates you are given are x and y coordinates, that is (x,y). plug the coordinates into the equation and see if the result is true, that is, see if it correctly solves the equation. If it is true, then it is a point on the graph of the equation. x+3y=13 (4,-3) 4+3(-3)=4-9=-5 this isnot 13 so NO (4,4) 4+4(-3)=4-12=-8 this is not 13 so NO (-5,6) -5+3(6)=-5+18=13 this is 13 so YES (-2,3) -2+3(3)=-2+9=7 this is not 13 so NO (3,2) 3+3(2)=3+6+9 this is not 13 so NO only C is on the graph
How can I find the coordinates of the point on y=x2 at which the gradient is 2?
The first thing to do is to differentiate [math]y=x^2 [/math]which gives you the gradient function.By inserting equating to 2, you find the value of [math]x[/math][math]dy/dx = 2x[/math][math]2x = 2[/math][math]x=1[/math]Plug in the value of [math]x[/math], back into the original function to find the value of [math]y[/math] that corresponds.[math]y=1^2[/math][math]y=1[/math]So the point where the gradient is 2, is [math](1,1)[/math]
What are the coordinates of the point D if A (-3, 1), B (4, 0), C (0, -3) and D are corners of a square?
Given,[math]A=(x_a,y_a)=(-3,1)[/math][math]B=(x_b,y_b)=(4,0)[/math][math]C=(x_c,y_c)=(0,-3)[/math]So we have,[math]AB =\sqrt{(x_a-x_b)^2+(y_a-y_b)^2}=\sqrt{50}[/math][math]BC =\sqrt{(x_b-x_c)^2+(y_b-y_c)^2}=5[/math][math]AC =\sqrt{(x_a-x_c)^2+(y_a-y_c)^2}=5[/math]We note that,[math]BC^2+AC^2=AB^2[/math] [because [math]5^2+5^2=50[/math]]If the three points are part of a square, then we can say that [math]AB[/math] is one of it's diagonal and [math]BC[/math] and [math]AC[/math] are two of it's equal sides.To complete a square, we need a fourth point which is at a distance [math]5[/math], from both [math]A[/math] and [math]B[/math]. There would be two such points. [math]C[/math] is one of them. We need to find the other one, point [math]D[/math] [math](x_d,y_d)[/math], such that,[math]AD=BD=5[/math][math]\sqrt{(x_a-x_d)^2+(y_a-y_d^2)}=\sqrt{(x_b-x_d)^2+(y_b-y_d)^2}=5[/math][math]\sqrt{(-3-x_d)^2+(1-y_d^2)}=\sqrt{(4-x_d)^2+(0-y_d)^2}=5[/math]Solving which gives us two solutions: [math](0, -3)[/math] and [math](1,4)[/math]. As expected, one of them is [math]C[/math]. The other one,[math]D=(x_d,y_d)=(1,4)[/math]
The polar coordinates of a point are (2, 319). What are the Cartesian coordinates of that point?
Did you mean to say that your point in polar coordinates is [math]2\measuredangle319^{\circ}[/math]? You can convert to Cartesian coordinates [math](x,y)[/math] from polar coordinates r[math]\measuredangle\theta[/math] using the following equations.[math]x = r \cos \theta[/math][math]y = r \sin \theta[/math]Using the numbers you gave leads to[math]x = 2 \cos \left( 319^{\circ} \right) = 1.5094[/math][math]y = 2 \sin \left( 319^{\circ} \right) = -1.3131[/math]So the final answer is[math]2\measuredangle319^{\circ} = (1.5094,-1.3131)[/math]
P and Q are the points with co-ordinates (2, -1) and (-3, 4). Can you find the coordinates of the point R such that PR is 2/5 of PQ?
Here, we are given the ratio PR/PQ, not PR/RQ. So, we first need to find the ratio PR/RQ. Let PR and PQ equal 2x and 5x, keeping in mind their ratio. So, RQ=PQ-PR=3x. Hence, the ratio PR/RQ=2/3. Now, we can use the section formula to determine the coordinates of R.If the division is internal, the coordinates of R can be determined as follows.If the division is external, the coordinates of R can be determined as follows.The position of the two points becomes evident from the following diagram, where the internal and external positions of R are shown.
Find the exact coordinates of the point at which the following curve is steepest?
The curve is steepest at the inflection point, where the curvature changes sign. This occurs when the second derivative is zero. y = 60 / [1 + 6e^(-2t)] y' = 720e^(2t) / [e^(2t) + 6]^2 y" = 1440e^(2t)[6 - e^(2t)] / [e^(2t) + 6]^3 y" = 0 implies e^(2t)[6 - e^(2t)] = 0 That is, when either - e^(2t) = 0, which is impossible, or 6 - e^(2t) = 0, which implies e^(2t) = 6, so 2t = ln(6). Thus, t = ln(6)/2 ≈ 0.8959.
The coordinates of a point A in the figure below are (a,b) where |a| > |2b|. Which of the following could be the slope of AB?
Let the angle AOX’ be [math]\theta[/math], then because [math]|a| > |2b|[/math] or [math]\dfrac{|b|}{|a|}< \dfrac{1}{2}[/math], we know that[math]tan \theta = \dfrac{|b|}{|a|} < \dfrac{1}{2} [/math]In this case [math]m = -tan \theta[/math], because slope is measured taking positive x-axis direction as base, (the angle for slope would be [math]\pi - \theta[/math])but [math]tan \theta < \dfrac{1}{2}[/math]or [math]-tan \theta > -\dfrac{1}{2}[/math]or [math]m > -\dfrac{1}{2}[/math]Because of the diagram , we can be sure that the slope is negative. Only one option satisfies the inequality and this condition, [math]m = -1[/math]So, ans is option a. [math]-1[/math]