Answer:
i think 105 is the correct answer Hope it helps you have a great day keep smiling be happy stay
Answer:
Step-by-step explanation:
(x+1)(x-3)(x-4) = (x+1)(x²-7x+12) = x³-6x²+5x+12
:::::
(x+3)(x-0)(x-0)(x-5) = (x+3)(x²)(x-5) = (x²-2x-15)x² = x⁴-2x³-15x²
:::::
(x+2)³ = x³ + 3·x²·2 + 3·x·2² + 2³ = x³ + 6x² + 12x + 8
:::::
(x+3)²(x+1) = (x²+6x+9)(x+1) = x³ + 7x² + 15x + 9
Answer:
Step-by-step explanation:
We are given that 
and 
are the zeroes of the polynomial 
We have to find a quadratic polynomial whose zeroes are 
and 
.
General quadratic equation
We get
Substitute the values
Hence, the quadratic polynomial whose zeroes are and is given by
Answer:
Hope this helps!
Please see through the steps below
(1/10 (x)) fewer than the sum of 1/5 (y) and 2x so
1/10 (x)=1/10 times x=x/10
1/5 (y)=y/5
-x/10+y/5+2x
simplified it is
19x/10+y/5
