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Cloud [144]
3 years ago
8

If it takes 16 men 15 days to complete a work,how long will it take 20 men?​

Mathematics
2 answers:
ASHA 777 [7]3 years ago
6 0

Answer:

10 days and 16 hours.

Step-by-step explanation:

15 men work in 16 days to complete the work. So 1 man contributes 16/15 days.  

For 20 men there is an addition of 5 men. [20–15)* 16/15] days = 16/3 days. Substract 16/3 days from 16 days = 16 - (16/3) = 10 days and 16 hours.

igor_vitrenko [27]3 years ago
5 0

Answer:

12

Step-by-step explanation:

Let 20 men will take x days to complete the work.

Since, there is a relation of inverse proportion between no of men and no of days.

Therefore,

20x = 16 \times 15 \\  \\ x =  \frac{240}{20}  \\  \\ x = 12

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Watermelons cost $0.35 per pound. Joseph’s watermelon costs between $4.20 and $5.25. Which compound inequality correctly represe
Ahat [919]
Working is attached :)

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3 years ago
First you to find the worksheet and download it<br> plase I need help
Goshia [24]

Answer:

a) The horizontal asymptote is y = 0

The y-intercept is (0, 9)

b) The horizontal asymptote is y = 0

The y-intercept is (0, 5)

c) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

d) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

e) The horizontal asymptote is y = -1

The y-intercept is (0, 7)

The x-intercept is (-3, 0)

f) The asymptote is y = 2

The y-intercept is (0, 6)

Step-by-step explanation:

a) f(x) = 3^{x + 2}

The asymptote is given as x → -∞, f(x) = 3^{x + 2} → 0

∴ The horizontal asymptote is f(x) = y = 0

The y-intercept is given when x = 0, we get;

f(x) = 3^{0 + 2} = 9

The y-intercept is f(x) = (0, 9)

b) f(x) = 5^{1  - x}

The asymptote is fx) = 0 as x → ∞

The asymptote is y = 0

Similar to question (1) above, the y-intercept is f(x) = 5^{1  - 0} = 5

The y-intercept is (0, 5)

c) f(x) = 3ˣ + 3

The asymptote is 3ˣ → 0 and f(x) → 3 as x → ∞

The asymptote is y = 3

The y-intercept is f(x) = 3⁰ + 3= 4

The y-intercept is (0, 4)

d) f(x) = 6⁻ˣ + 3

The asymptote is 6⁻ˣ → 0 and f(x) → 3 as x → ∞

The horizontal asymptote is y = 3

The y-intercept is f(x) = 6⁻⁰ + 3 = 4

The y-intercept is (0, 4)

e) f(x) = 2^{x + 3} - 1

The asymptote is 2^{x + 3}  → 0 and f(x) → -1 as x → -∞

The horizontal asymptote is y = -1

The y-intercept is f(x) =  2^{0 + 3} - 1 = 7

The y-intercept is (0, 7)

When f(x) = 0, 2^{x + 3} - 1 = 0

2^{x + 3} = 1

x + 3 = 0, x = -3

The x-intercept is (-3, 0)

f) f(x) = \left (\dfrac{1}{2} \right)^{x - 2} + 2

The asymptote is \left (\dfrac{1}{2} \right)^{x - 2} → 0 and f(x) → 2 as x → ∞

The asymptote is y = 2

The y-intercept is f(x) = f(0) = \left (\dfrac{1}{2} \right)^{0 - 2} + 2 = 6

The y-intercept is (0, 6)

7 0
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Since the radius is 12, the equation is
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