I will suppose that your queation is 10<x<y<14
there are only two odd between 10 and 14
which are 11 and 13
since x is smaller than y
x=11 and y=13
there sum will be x+y=11+13=24
D≈2π(6371-1737.5)
d≈9267π
d≈29113.1km (to nearest tenth of a kilometer)
PART A:
The given quadratic equation is 2x²-10x-8=0
The radicand is given by b²-4ac where a, b, and c are the constants in a quadratic form ax²+bx+c
From the given equation, we have
a = 2
b = -10
c = -8
Radicand b²-4ac = (-10)² - 4(2)(-8) = 100 + 64 = 164
The radicand is >0 hence the quadratic equation has two distinct roots
PART B:
4x²-12x+5 = 0
We can use the factorization method to solve the equation
Firstly, we multiply 4 by 5 to get 20
Then we find the pair of numbers that multiply gives 20 and sum gives -12
The pair of number is -2 and -10
Rewriting the equation
4x²-2x-10x+5 = 0
2x(2x-1)-5(2x-1) = 0
(2x-1)(2x-5) = 0
2x-1 = 0 and 2x-5 = 0
x = 1/2 and x = 5/2
More information is needed to answer this question.
Answer:
Step-by-step explanation:
I attached the graphs
