Y - x = 154
let z = xy = x(154 + x) = x^2 + 154x
we need to find the minimum value of z
z' = 2x + 154 which = zero for minm or maxm value
x = -77 and y = 154 -77 = 77
so the 2 numbers are -77 and 77
The simplified version of this expression is 110g^4h^16v^6. first, remove the parenthesis. the expression becomes 10g^3h^8v^6 times 11gh^8. then, calculate the product.
C^2=a^2+b^2 we are given the hypotenuse of 19 units and one leg of 8 units so:
19^2=8^2+b^2
361=64+b^2
b^2=297
b=√297
b≈17.2 to the nearest tenth
Answer:
reflection
Step-by-step explanation:
think of it as looking in a mirror
