Answer:
Step-by-step explanation:
(-3,4) and (5,1)
(-3,4) is in quadrant 2 and (5,1) is in quadrant 1. So think of that.
(x+8, y-3)
8 units to the right and 3 units down
The trigonometric function is given as
Apply the half angle identity to find the value of tan 75 ,
Here,
![\tan (75^{\circ})=\frac{\frac{1}{2}}{1-\frac{\sqrt[]{3}}{2}}=\frac{\frac{1}{2}}{\frac{2-\sqrt[]{3}}{2}}^{}](https://tex.z-dn.net/?f=%5Ctan%20%2875%5E%7B%5Ccirc%7D%29%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%7D%7B1-%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%7D%3D%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%7D%7B%5Cfrac%7B2-%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%7D%5E%7B%7D)
![\tan 75^{\circ}=\frac{1}{2-\sqrt[]{3}}](https://tex.z-dn.net/?f=%5Ctan%2075%5E%7B%5Ccirc%7D%3D%5Cfrac%7B1%7D%7B2-%5Csqrt%5B%5D%7B3%7D%7D)
Now rationalize the function.
![\tan 75^{\circ}=\frac{1}{2-\sqrt[]{3}}\times\frac{2+\sqrt[]{3}}{2+\sqrt[]{3}}=\frac{2+\sqrt[]{3}}{4-3}=\frac{2+\sqrt[]{3}}{1}](https://tex.z-dn.net/?f=%5Ctan%2075%5E%7B%5Ccirc%7D%3D%5Cfrac%7B1%7D%7B2-%5Csqrt%5B%5D%7B3%7D%7D%5Ctimes%5Cfrac%7B2%2B%5Csqrt%5B%5D%7B3%7D%7D%7B2%2B%5Csqrt%5B%5D%7B3%7D%7D%3D%5Cfrac%7B2%2B%5Csqrt%5B%5D%7B3%7D%7D%7B4-3%7D%3D%5Cfrac%7B2%2B%5Csqrt%5B%5D%7B3%7D%7D%7B1%7D)
Again simplify the trigonometric function,
Hence the answer is 3.732.
Well, I would first make a table and then I would find the slope. That is my advice.
<span>Vector Equation
(Line)</span>(x,y) = (x,y) + t(a,b);tERParametric Formx = x + t(a), y = y + t(b); tERr = (-4,-2) + t((-3,5);tERFind the vector equation of the line passing through A(-4,-2) & parallel to m = (-3,5)<span>Point: (2,5)
Create a direction vector: AB = (-1 - 2, 4 - 5)
= (-3,-1) or (3,1)when -1 (or any scalar multiple) is divided out.
r = (2,5) + t(-3,-1);tER</span>Find the vector equation of the line passing through A(2,5) & B(-1,4)<span>x = 4 - 3t
y = -2 + 5t
;tER</span>Write the parametric equations of the line passing through the line passing through the point A(4,-2) & with a direction vector of m =(-3,5)<span>Create Vector Equation first:
AB = (2,8)
Point: (4,-3)
r = (4,-3) + (2,8); tER
x = 4 + 2t
y = -3 + 8t
;tER</span>Write the parametric equations of the line through A(4,-3) & B(6,5)<span>Make parametric equations:
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in -3
-3 = 5 + 4t
(-8 - 5)/4 = t
-2 = t
For y sub in -8
-8 = -2 + 3t
(-8 + 2)/3 = t
-2 = t
Parameter 't' is consistent so pt(-3,-8) is on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (-3,-8) on the line?<span>Make parametric equations:
x = 5 + 4t
y = -2 + 3t ; tER
For x sub in 1
-1 = 5 + 4t
(-1 - 5)/4 = t
-1 = t
For y sub in -7
-7 = -2 + 3t
(-7 + 2)/3 = t
-5/3 = t
Parameter 't' is inconsistent so pt(1,-7) is not on the line.</span>Given the equation r = (5,-2) + t(4,3);tER, is (1,-7) on the line?<span>Use parametric equations when generating points:
x = 5 + 4t
y = -2 + 3t ;tER
X-int:
sub in y = 0
0 = -2 + 3t
solve for t
2/3 = t (this is the parameter that will generate the x-int)
Sub t = 2/3 into x = 5 + 4t
x = 5 + 4(2/3)
x = 5 + (8/3)
x = 15/3 + (8/3)
x = 23/3
The x-int is (23/3, 0)</span>What is the x-int of the line r = (5,-2) + t(4,3); tER?Note: if they define the same line: 1) Are their direction vectors scalar multiples? 2) Check the point of one equation in the other equation (LS = RS if point is subbed in)What are the two requirements for 2 lines to define the same line?
9514 1404 393
Answer:
k = 7
m = 8
Step-by-step explanation:
The diagonals bisect each other, so each half is equal to the opposite half.
k +4 = 11 . . . . down-sloping diagonal
k = 7
__
m = 8 . . . . up-sloping diagonal
