Suppose the number of water lilies in pond doubles every day. If the pond was half covered on Monday, when was it only one-forth
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<u>Annotation</u>
General formula for distance-time-velocity relationship is as following
d = v × t
The velocity of the first car will be v₁, the time is 2 hours, the distance will be d₁.
The velocity of the second car will be v₂, the time is 2 hours, the distance will be d₂.
One of them traveling 5 miles per hour faster than the others. That means the velocity of the first car is 5 miles per hour more than the velocity of the second car.
v₁ = v₂ + 5 (first equation)
The distance of the two cars after two hours will be 262 miles apart. Because they go to opposite direction, we could write it as below.
d₁ + d₂ = 262 (second equation)
Plug the d-v-t relationship to the second equation
d₁ + d₂ = 262
v₁ × t + v₂ × t = 262
v₁ × 2 + v₂ × 2 = 262
2v₁ + 2v₂ = 262
Plug the v₁ as (v₂+5) from the first equation
2v₁ + 2v₂ = 262
2(v₂ + 5) + 2v₂ = 262
2v₂ + 10 + 2v₂ = 262
4v₂ + 10 = 262
4v₂ = 252
v₂ = 252/4
v₂ = 63
The second car is 63 mph fast.
Find the velocity of the first car, use the first equation
v₁ = v₂ + 5
v₁ = 63 + 5
v₁ = 68
The first car is 68 mph fast.
Answer
General formula for distance-time-velocity relationship is as following
d = v × t
The velocity of the first car will be v₁, the time is 2 hours, the distance will be d₁.
The velocity of the second car will be v₂, the time is 2 hours, the distance will be d₂.
One of them traveling 5 miles per hour faster than the others. That means the velocity of the first car is 5 miles per hour more than the velocity of the second car.
v₁ = v₂ + 5 (first equation)
The distance of the two cars after two hours will be 262 miles apart. Because they go to opposite direction, we could write it as below.
d₁ + d₂ = 262 (second equation)
Plug the d-v-t relationship to the second equation
d₁ + d₂ = 262
v₁ × t + v₂ × t = 262
v₁ × 2 + v₂ × 2 = 262
2v₁ + 2v₂ = 262
Plug the v₁ as (v₂+5) from the first equation
2v₁ + 2v₂ = 262
2(v₂ + 5) + 2v₂ = 262
2v₂ + 10 + 2v₂ = 262
4v₂ + 10 = 262
4v₂ = 252
v₂ = 252/4
v₂ = 63
The second car is 63 mph fast.
Find the velocity of the first car, use the first equation
v₁ = v₂ + 5
v₁ = 63 + 5
v₁ = 68
The first car is 68 mph fast.
Answer
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)