60x1.15=69
Retail price is $69
Answer:
Therefore the required polynomial is
M(x)=0.83(x³+4x²+16x+64)
Step-by-step explanation:
Given that M is a polynomial of degree 3.
So, it has three zeros.
Let the polynomial be
M(x) =a(x-p)(x-q)(x-r)
The two zeros of the polynomial are -4 and 4i.
Since 4i is a complex number. Then the conjugate of 4i is also a zero of the polynomial i.e -4i.
Then,
M(x)= a{x-(-4)}(x-4i){x-(-4i)}
=a(x+4)(x-4i)(x+4i)
=a(x+4){x²-(4i)²} [ applying the formula (a+b)(a-b)=a²-b²]
=a(x+4)(x²-16i²)
=a(x+4)(x²+16) [∵i² = -1]
=a(x³+4x²+16x+64)
Again given that M(0)= 53.12 . Putting x=0 in the polynomial
53.12 =a(0+4.0+16.0+64)

=0.83
Therefore the required polynomial is
M(x)=0.83(x³+4x²+16x+64)
Answer: No solutions
If i put any combo of horses of single or odd, always an even number at last stall.
Combine like terms.
Since 7x and 6x both have a variable of x, you can combine them.

After that, you are left with:

You also have two numbers without variables; 11 and 3. You can also combine those.

So, your final answer would be:

Answer:
D. Section A; students in this section scored between 1 and 10
Step-by-step explanation:
This answer is correct because the graph shows everything that was explained in the answer. Hope that helps!!