Answer: 4/5 17/20 9/10
Step-by-step explanation:
Find a common denominator
4/5-16/20
17/20-17/20
9/10-18/20
Step-by-step explanation:
(1) Factor the GCF out of the trinomial on the left side of the equation. (2 points: 1 for the GCF, 1 for the trinomial) 2x²+6x-20=0
2x²+6x-20
2(x²+3x-10)
the factors are 2 and (x²+3x-10)
(2) Factor the polynomial completely. (4 points: 2 point for each factor)
2(x²+3x-10)
2(x²-2x+5x-10)
2(x(x-2) + 5(x-2)) group like terms
2(x+5)(x-2)
(3) If a product is equal to zero, we know at least one of the factors must be zero. And the constant factor cannot be zero. So set each binomial factor equal to 0 and solve for x, the width of your project. (2 points: 1 point for each factor)
constant = 2 cannot be zero
the other factors are (x+5) and (x-2)
(x+5)=0 => x= -5
or
(x-2)=0 => x=2
(4) What are the dimensions of your project? Remember that the width of your project is represented by x. (2 points: 1 point for each dimension)
thank you so much, sorry if it's a little confusing!!
(it is indeed confusing, because physical dimensions cannot be negative)
The dimensions of the project (assumed a rectangle) are +2 and -5
As stated in (i), the equation of the line is: ln y = -0.015x + .26
(By the way, I checked your answers for parts (i) and (ii) and they are both correct)
(iii)
Plug in (1.1) for y and solve:
ln (1.1) = -0.015x + .26
0.095 = -0.015x + .26
-0.165 = - 0.015x
10.979 = x
Answer: x = 10.979
Answer:
- P(t) = 100·2.3^t
- 529 after 2 hours
- 441 per hour, rate of growth at 2 hours
- 5.5 hours to reach 10,000
Step-by-step explanation:
It often works well to write an exponential expression as ...
value = (initial value)×(growth factor)^(t/(growth period))
(a) Here, the growth factor for the bacteria is given as 230/100 = 2.3 in a period of 1 hour. The initial number is 100, so we can write the pupulation function as ...
P(t) = 100·2.3^t
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(b) P(2) = 100·2.3^2 = 529 . . . number after 2 hours
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(c) P'(t) = ln(2.3)P(t) ≈ 83.2909·2.3^t
P'(2) = 83.2909·2.3^2 ≈ 441 . . . bacteria per hour
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(d) We want to find t such that ...
P(t) = 10000
100·2.3^t = 10000 . . . substitute for P(t)
2.3^t = 100 . . . . . . . . divide by 100
t·log(2.3) = log(100)
t = 2/log(2.3) ≈ 5.5 . . . hours until the population reaches 10,000
Answer:
(-1.5, 0) and (0, -6)
Step-by-step explanation:
