Well, parallel lines have the same exact slope, so hmmm what's the slope of the one that runs through <span>(0, −3) and (2, 3)?
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so, we're really looking for a line whose slope is 3, and runs through -1, -1
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![\bf \begin{array}{ccccccccc} &&x_1&&y_1\\ % (a,b) &&(~ -1 &,& -1~) \end{array} \\\\\\ % slope = m slope = m\implies 3 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-1)=3[x-(-1)] \\\\\\ y+1=3(x+1)](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%20-1%20%26%2C%26%20-1~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20slope%20%20%3D%20m%0Aslope%20%3D%20%20m%5Cimplies%203%0A%5C%5C%5C%5C%5C%5C%0A%25%20point-slope%20intercept%0A%5Cstackrel%7B%5Ctextit%7Bpoint-slope%20form%7D%7D%7By-%20y_1%3D%20m%28x-%20x_1%29%7D%5Cimplies%20y-%28-1%29%3D3%5Bx-%28-1%29%5D%0A%5C%5C%5C%5C%5C%5C%0Ay%2B1%3D3%28x%2B1%29)
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Uhhhhhhhhhh i don’t really know
Answer:
2X/(3+X)
Step-by-step explanation:
X = 3m/2-m
Cross multiply both sides
X × (2-m)= 3m
2X-Xm= 3m
2X= 3m+Xm
2X= m(3+X)
Divide both sides by the coefficient of m which is (3+X)
m= 2X/(3+X)
Answer:
71
Step-by-step explanation:
[6x4(15/5)]+[2^2+(1x-5)]
BIDMAS (aka PEDMAS etc.)
Brackets
Indices
Division/Multiplication
Addition/Subtraction
[6 x 4(15/5)] + [2^2 + (1x-5)]
[6 x 4(3)] + [2^2 + (-5)]
[6 x 4(3)] + [4 + -5]
[6 x 12] + [4 + -5]
[72] + [4 + -5]
72 + [-1]
72 + - 1
72 - 1
71
