Answer: Yes, the point (3,4) is a solution to the system.
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Proof of this:
Replace x with 3 and y with 4 in the first equation
x+y = 7
3+4 = 7
7 = 7
This confirms the first equation. Repeat for the second equation
x-2y = -5
3-2(4) = -5
3 - 8 = -5
-5 = -5
We get true equations for both when we plug in (x,y) = (3,4). This confirms it is a valid solution to the system of equations. It turns out it's the only solution to this system of equations. Visually, the two lines cross at the single location (3,4).
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)
It’s 5,905 bc you need to add all of you distilled water to know if you have enough