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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Answer:
there are 48cubes fetted into a box
To find Angle A we use cosine
cos ∅ = adjacent / hypotenuse
From the question
The adjacent is 17
The hypotenuse is 38
So we have
cos A = 17/38
A = cos-¹ 17/38
A = 63.4
<h3>A = 63° to the nearest degree</h3>
To find Angle C we use sine
sin ∅ = opposite / hypotenuse
From the question
The opposite is 17
The hypotenuse is 38
So we have
sin C = 17/38
C = sin-¹ 17/38
C = 26.57
<h3>C = 27° to the nearest degree</h3>
Hope this helps you
5(x^2 - 14x + 258/5)
= 5((x-7)^2 -49+258/5)
= 5((x-7)^2 + 13/5)
= 5(x-7)^2 + 13
I belive this is how it should be done
