Write a power represented with a positive base and a positive exponent whose value is less than the base
1 answer:
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Answer:
0
1
Step-by-step explanation:
First question:
You are given a side, a, and its opposite angle, A. You are also given side b. Use that in the law of sines and solve for the other angle, B.
The sine function can never equal 2, so there is no triangle in this case.
Answer: no triangle
Second question:
You are given a side, b, and its opposite angle, B. You are also given side c. Use that in the law of sines and solve for the other angle, C.
One triangle exists for sure. Now we see if there is a second one.
Now we look at the supplement of angle C.
m<C = 52.5°
supplement of angle C: m<C' = 180° - 52.5° = 127.5°
We add the measures of angles B and the supplement of angle C:
m<B + m<C' = 63° + 127.5° = 190.5°
Since the sum of the measures of these two angles is already more than 180°, the supplement of angle C cannot be an angle of the triangle.
Answer: one triangle
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)