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Darina [25.2K]
3 years ago
11

For the function y=12x+6, if the input is x=−4, the output is y=

Mathematics
2 answers:
Sav [38]3 years ago
3 0

Answer:

12(4) +6

48+6 = 54

y=54

Step-by-step explanation:

melamori03 [73]3 years ago
3 0

Step-by-step explanation:

Given function: y = 12x + 6

Input: x = -4

Replacing the value of x in function y,

y = 12 x (-4) + 6

y = -48 + 6

y = -42

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A house purchased 5 years ago for 100,000 was sold for 161,051. Assuming exponential growth , approximate the annual growth rate
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8 0
2 years ago
Prove that the Greatest Integer Function f: R -> R given by f(x) = [x], is neither one-once nor onto, where [x] denotes the g
Karo-lina-s [1.5K]
F: R → R is given by, f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1
So, f(1.2) = f(1.9), but 1.2 ≠ 1.9
f is not one-one

Now, consider 0.7 ε R
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ε R such that f(x) = 0.7

So, f is not onto
Hence, the greatest integer function is neither one-one nor onto.

The answer was quite complicated but I hope it will help you.
8 0
3 years ago
MCR3U1 Culminating 2021.pdf
11111nata11111 [884]

Answer:

(a) y = 350,000 \times (1 + 0.07132)^t

(b) (i) The population after 8 hours is 607,325

(ii) The population after 24 hours is 1,828,643

(c) The rate of increase of the population as a percentage per hour is 7.132%

(d) The doubling time of the population is approximately, 10.06 hours

Step-by-step explanation:

(a) The initial population of the bacteria, y₁ = a = 350,000

The time the colony grows, t = 12 hours

The final population of bacteria in the colony, y₂ = 800,000

The exponential growth model, can be written as follows;

y = a \cdot (1 + r)^t

Plugging in the values, we get;

800,000 = 350,000 \times (1 + r)^{12}

Therefore;

(1 + r)¹² = 800,000/350,000 = 16/7

12·㏑(1 + r) = ㏑(16/7)

㏑(1 + r) = (㏑(16/7))/12

r = e^((㏑(16/7))/12) - 1 ≈ 0.07132

The  model is therefore;

y = 350,000 \times (1 + 0.07132)^t

(b) (i) The population after 8 hours is given as follows;

y = 350,000 × (1 + 0.07132)⁸ ≈ 607,325.82

By rounding down, we have;

The population after 8 hours, y = 607,325

(ii) The population after 24 hours is given as follows;

y = 350,000 × (1 + 0.07132)²⁴ ≈ 1,828,643.92571

By rounding down, we have;

The population after 24 hours, y = 1,828,643

(c) The rate of increase of the population as a percentage per hour =  r × 100

∴   The rate of increase of the population as a percentage = 0.07132 × 100 = 7.132%

(d) The doubling time of the population is the time it takes the population to double, which is given as follows;

Initial population = y

Final population = 2·y

The doubling time of the population is therefore;

2 \cdot y = y \times (1 + 0.07132)^t

Therefore, we have;

2·y/y =2 = (1 + 0.07132)^t

t = ln2/(ln(1 + 0.07132)) ≈ 10.06

The doubling time of the population is approximately, 10.06 hours.

8 0
3 years ago
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