let's recall that in a Kite the diagonals meet each other at 90° angles, Check the picture below, so we're looking for the equation of a line that's perpendicular to BD and that passes through (-1 , 3).
keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of BD
so we're really looking for the equation of a line whose slope is -1/3 and passes through point A
![(\stackrel{x_1}{-1}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{-\cfrac{1}{3}}[x-\stackrel{x_1}{(-1)}]\implies y-3=-\cfrac{1}{3}(x+1) \\\\\\ y-3=-\cfrac{1}{3}x-\cfrac{1}{3}\implies y=-\cfrac{1}{3}x-\cfrac{1}{3}+3\implies y=-\cfrac{1}{3}x+\cfrac{8}{3}](https://tex.z-dn.net/?f=%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B3%7D%29%5Cqquad%20%5Cqquad%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20-%5Ccfrac%7B1%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-%5Cstackrel%7By_1%7D%7B3%7D%3D%5Cstackrel%7Bm%7D%7B-%5Ccfrac%7B1%7D%7B3%7D%7D%5Bx-%5Cstackrel%7Bx_1%7D%7B%28-1%29%7D%5D%5Cimplies%20y-3%3D-%5Ccfrac%7B1%7D%7B3%7D%28x%2B1%29%20%5C%5C%5C%5C%5C%5C%20y-3%3D-%5Ccfrac%7B1%7D%7B3%7Dx-%5Ccfrac%7B1%7D%7B3%7D%5Cimplies%20y%3D-%5Ccfrac%7B1%7D%7B3%7Dx-%5Ccfrac%7B1%7D%7B3%7D%2B3%5Cimplies%20y%3D-%5Ccfrac%7B1%7D%7B3%7Dx%2B%5Ccfrac%7B8%7D%7B3%7D)
8.77 x 10^1 i believe but would double check
Answer: A
Explanation:
(6x^2 - x + 8) - (x^2 + 2)
6x^2 - x + 8 - x^2 - 2
= 5x^2 - x + 6
Answer:
Below.
Step-by-step explanation:
f) (a + b)^3 - 4(a + b)^2
The (a+ b)^2 can be taken out to give:
= (a + b)^2(a + b - 4)
= (a + b)(a + b)(a + b - 4).
g) 3x(x - y) - 6(-x + y)
= 3x( x - y) + 6(x - y)
= (3x + 6)(x - y)
= 3(x + 2)(x - y).
h) (6a - 5b)(c - d) + (3a + 4b)(d - c)
= (6a - 5b)(c - d) + (-3a - 4b)(c - d)
= -(c - d)(6a - 5b)(3a + 4b).
i) -3d(-9a - 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b).
= (3d + 2c)(9a + 2b).
j) a^2b^3(2a + 1) - 6ab^2(-1 - 2a)
= a^2b^3(2a + 1) + 6ab^2(2a + 1)
= (2a + 1)( a^2b^3 + 6ab^2)
The GCF of a^2b^3 and 6ab^2 is ab^2, so we have:
(2a + 1)ab^2(ab + 6)
= ab^2(ab + 6)(2a + 1).
Answer:
Rectangle ABCDABCDA, B, C, D is graphed in the coordinate plane. vertices of the rectangle: A(-1, -6),A(−1,−6),A, left parenthesis, minus, 1, comma, minus, 24+2 ps y+4 sr y+5 qR 2x + 3 pq 2y+2.
Step-by-step explanation:
