X+y+5=0
y=-x-5
If a solution exists y=y so we can say
x^2-9x+10=-x-5 add x+5 to both sides
x^2-8x+15=0 now factor
x^2-3x-5x+15=0
x(x-3)-5(x-3)
(x-5)(x-3) so x=3 and 5, using y=-x-5
y(3)=-8 and y(5)=-10
So the two solutions are:
(3,-8) and (5,-10)
Answer:
7 1/17
Step-by-step explanation:
A figure can be helpful.
The inscribed semicircle has its center at the midpoint of th base. It is tangent to the side of the isosceles triangle, so a radius makes a 90° angle there.
The long side of the isosceles triangle can be found from the Pythagorean theorem to be ...
BC² = BD² +CD²
BC² = 8² +15² = 289
BC = √289 = 17
The radius mentioned (DE) creates right triangles that are similar to ∆BCD. In particular, we have ...
(long side)/(hypotenuse) = DE/BD = CD/BC
DE = BD·CD/BC = 8·15/17
DE = 7 1/17 ≈ 7.059
Answer:
Statements: 6, 1, 8, 3, 4, 7
Step-by-step explanation:
y = 3x/(8 + x)
x = 3y/(8 + y)
x(8 + y) = 3y
8x + xy = 3y
8x = 3y - xy
8x = y(3 - x)
y = 8x/(3 - x)
y = f^-(x) = 8x/(3 - x)
Sun is 35. Since=cos(90-x)