This gives you three simultaneous equations:
6 = a + c
7 = 4a + c
1 = c
<u>c = 1
</u><u /><u />
If c =1,
6 = a + 1
<u>a = 5
</u><u /><u />
This doesn't work in the second equation, so the quadratic that goes through these points is not in the form y = ax^2 + bx + c
Was there supposed to be a b in the equation?
4z>-60
divideby 4
z>-15
x+19 <=-5
subtract 19 from each side
x<= -24
Y = mx + c, where m is your gradient and c is your y intercept.
If the line is perpendicular to the line y = 3x - 5, the line's gradient = -(1/3) as when it comes to perpendicular bisector, m1/m2 = -1
y = -(1/3)x + C
To find C sub in x=2 and y=2 as it goes through (2,2)
2 = -(1/3)2 + C
2 = -(2/3) + C
2 + 2/3 = C
C = 8/3
There your equation of a line is
y = -(1/3)x + 8/3
Answer:
b. (√15)/4
Step-by-step explanation:
Since Sin Ф = (opposite side)/Hypotenuse, we have 2 sides of a right triangle.
Use Pythagorean theorem to solve for the missing leg (the adjacent side)
1² + b² = 4²
1 + b² = 16
b² = 15
b = √15
So the adjacent side is √15, so Cos Ф = (√15)/4
