Answer:
Average rate of change = -6
Step-by-step explanation:
The average rate of change over the interval (a,b) is given by;
[ f(b) - f(a)] / (b-a)........................where interval is (a,b)
(a,b) interval =(-3,1)
Where;
f(x)= -12 (x+2)² +5
I will mark them as brainliest who solve it
1 answer:
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Answer:
AKPOS/ACGF = 28√3 -48 ≈ 0.497423
Step-by-step explanation:
Let's assume the square is a unit square. Then the height of the triangle is ...
1 +(√3)/2 = h
and half the base of the triangle is ...
h/√3 = (√3)/3 +1/2 = b/2
The area of the triangle is the product of these:
A = (1/2)bh
= (2 +√3)/2 × (3 +2√3)/6
= (6 +3√3 +4√3 +6)/12 = (12 +7√3)/12
So, the ratio of the area of the square to that of the triangle is the inverse of this, or ...
(square area)/(triangle area) = 12/(12+7√3)
(square area)/(triangle area) = 28√3 -48
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)