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3 years ago

A leaky faucet in kali's house drips water at a rate of 42 gallons per week. How many gallons does the faucet drip in three days

Mathematics

1 answer:

Ratling [72]3 years ago

5 0

Amount of water that the faucet leaks in 1 week = 42 gallons

That means Amount of water that the faucet in Kali's house leaks in 7 days = 42 gallons So The amount of water that the faucet in Kali's house leaks in 1 day = (42/7) gallons = 6 gallons Then The amount of water that the faucet in Kali's house leaks in 3 days = 6 * 3 gallons = 18 gallons

So it can said from the above deduction that the amount of water leaked by the faucet in Kali's house is 18 gallons.

hope this help .-.

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Which of the following expressions is equivalent to 14xy – 28x – 36y + 48? 2(7x – 6)(y – 2)(3y + 4) 2(7x – 6)(y – 2)(3y – 4) 2[7

Zolol [24]

The expression that is equivalent to 14xy – 28x – 36y + 48 is 2[7x(y-2)-6(3y-4)]

<h3>Factorizing expressions</h3>

Factorization is a way of separating the equations into two separate factors.

Given the expression below;

14xy – 28x – 36y + 48

Group

(14xy – 28x) – (36y + 48)

14x(y - 2) - 12(3y-4)

Factor out the value of 2 from both terms

2[7x(y-2)-6(3y-4)]

Hence the expression that is equivalent to 14xy – 28x – 36y + 48 is 2[7x(y-2)-6(3y-4)]

Learn more on factorization here: brainly.com/question/25829061

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2 years ago

A lampshade has a height of 12cm and upper and lower diameters of 10cm and 20cm A)what area of material is required to cover tha

netineya [11]

Answer:

$(a)\ Area = 195\pi$

$(b)\ Volume = 700\pi$

Step-by-step explanation:

Given

$h = 12$

$d_1 = 10$ --- lower diameter

$d_2 = 20$ --- upper diameter

Solving (a): The curved surface area

This is calculated as:

$Area = \pi l(r_1 + r_2)$

Where

$r_1 = 0.5 * d_1 = 0.5 * 10 = 5$ --- lower radius

$r_2 = 0.5 * d_2 = 0.5 * 20 = 10$ --- upper radius

And

$l = \sqrt{h^2 + (r_1 - r_2)^2$ ---- l represents the slant height of the frustrum

$l = \sqrt{12^2 + (5 - 10)^2$

$l = \sqrt{12^2 + (-5)^2$

$l = \sqrt{144 + 25$

$l = \sqrt{169$

$l = 13$

So, we have:

$Area = \pi l(r_1 + r_2)$

$Area = \pi * 13(5 + 10)$

$Area = \pi * 13(15)$

$Area = 195\pi$

Solving (b): The volume

This is calculated as:

$Volume = \frac{1}{3}\pi * h * (r_1^2 + r_2^2 + (r_1 * r_2))$

This gives:

$Volume = \frac{1}{3}\pi * 12 * (5^2 + 10^2 + (5 * 10))$

$Volume = \frac{1}{3}\pi * 12 * (25 + 100 + (50))$

$Volume = \frac{1}{3}\pi * 12 * (175)$

$Volume = \pi *4 * 175$

$Volume = 700\pi$

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