(x-3)^2 x=a number in this case
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)
1:3 can be another ratio of black fish to white fish
Answer:
19.2
Step-by-step explanation:
<u>1st Case:</u>
4 and 5 are legs of the right triangle.
Using the pythagorean therom: a^2+b^2=c^2
We can say that 4^2+5^2=x^2
16+25=x^2
41=x^2
x=√41
√41 is about 6.4
x=6.4
<u>2nd Case</u>
5 is the hypotenuse of the right triangle and 4 is the legs.
Using the pythagorean therom: a^2+b^2=c^2
We can say that 4^2+x^2=5^2
16+x^2=25
x^2=9
x=3
<u>Final Step</u>
We need to multiply the two possible lengths for x. So for case 1 the length of x was 6.4 and for case two the length was 3. 6.4*3=19.2
Anwser: <u>19.2</u>