The answer to this question is B) 0
First, let's re-arrange to slope-intercept form.
x + 8y = 27
Subtract 'x' to both sides:
8y = -x + 27
Divide 8 to both sides:
y = -1/8x + 3.375
So the slope of this line is -1/8, to find the slope that is perpendicular to this, we multiply it by -1 and flip it. -1/8 * -1 = 1/8, flipping it will give us 8/1 or 8.
So the slope of the perpendicular line will be 8.
Now we can plug this into point-slope form along with the point given.
y - y1 = m(x - x1)
y - 5 = 8(x + 5)
y - 5 = 8x + 40
y = 8x + 45
Take the 4cm side and the 6cm side.
If you hook them together at one end, then there's no way they can reach the ends of the 11cm side. The farthest they can reach is 10cm.
That's what choice-B says.
Given the equation - x² + 5x = 3, which can be rewritten as:
- x² + 5x - 3 = 0
where a = -1, b = 5 and c = -3.
Quadratic formula:
![\frac{-b\text{ }\pm\text{ }\sqrt[]{b^2\text{ - 4ac}}}{2a}](https://tex.z-dn.net/?f=%5Cfrac%7B-b%5Ctext%7B%20%7D%5Cpm%5Ctext%7B%20%7D%5Csqrt%5B%5D%7Bb%5E2%5Ctext%7B%20-%204ac%7D%7D%7D%7B2a%7D)
Now, we just replace the values of a, b and c on the equation above.
![\frac{-5\text{ }\pm\text{ }\sqrt[]{5^2\text{ - 4(-1)(3)}}}{2(-1)}](https://tex.z-dn.net/?f=%5Cfrac%7B-5%5Ctext%7B%20%7D%5Cpm%5Ctext%7B%20%7D%5Csqrt%5B%5D%7B5%5E2%5Ctext%7B%20-%204%28-1%29%283%29%7D%7D%7D%7B2%28-1%29%7D)
=
![\frac{5}{2}\text{ }\pm\text{ }\frac{\sqrt[]{13}}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B2%7D%5Ctext%7B%20%7D%5Cpm%5Ctext%7B%20%7D%5Cfrac%7B%5Csqrt%5B%5D%7B13%7D%7D%7B2%7D)
