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So hmm notice the picture below
you have the center, and a point on the circle... all you need is the radius
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 1}}\quad ,&{{ -2}})\quad % (c,d) &({{ 3}}\quad ,&{{ -4}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\leftarrow r](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%28%7B%7B%201%7D%7D%5Cquad%20%2C%26%7B%7B%20-2%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%26%28%7B%7B%203%7D%7D%5Cquad%20%2C%26%7B%7B%20-4%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%5Cleftarrow%20%20r)
then use that radius in the circle's equation
Answer:
1) x(x+7)
3) (x+5)*(x+4)
4) (x-6)*(x-8)
5) x(2x+21)+11
6) 5(a-5)*(a+5)
7) (4x+3)*(2x-3)
8) 4( x²-6xy+8)
Step-by-step explanation:
Not entirely sure, but it seem the answer could be B.) It will not be spread out vertically across the entire coordinate plane because in step 5, Nancy selected an incorrect scale on the y-axis.
Answer:
- 5/48, 3/16, .5, .75, 13
- 1/5, .35, 12/25, .5, 4/5
- -3/4, -7/10, 3/40, 8/10
- -.65, -3/8, 5/16, 2/4
Step-by-step explanation:
- 5/48 = 1.0291666666 | 3/16 = .1875 the rest is obvious
- 1/5 = .2 | 12/25 = .48 | 4/5 = .8
- -3/4 = -.75 | -7/10 = -.7 | 3/40 = .075 | 8/10 = .8
- -3/8 = -.375 | 5/16 = .3125 | 2/4 = .5