All categories

2 years ago

Easy question:

Mathematics

1 answer:

Delvig [45]2 years ago

8 0

Answer:

3 hours and 45 minutes

Step-by-step explanation:

let's assume the tank has a volume of x liters.

let's assume the speed of filling it is x/6 liters per hour for A and x/10 liters per hour for B.

the formula to calculate the time is:

time = volume / speed

If the pumps work together, the total speed is x/6 + x/10, which is 16/60 x

So the time this takes is:

time = x / (16/60 x) = 60/16 hours = 3.75 hours = 3 hours and 45 minutes.

You might be interested in

Hey, me again I NEED HELP ASAP!

LenKa [72]

THe answer is 36 BOI

3 0

2 years ago

A pet store has 11 puppies, including 4 poodles, 5 terriers, and 2 retrievers. If Rebecka and Aaron, in that order, each se

Nataliya [291]

The probability that Rebecka selects a terrier and Aaron selects a retriever is 10/121

<h3>How to determine the probability?</h3>

The distribution of the puppies is given as:

4 poodles, 5 terriers, and 2 retrievers

The probability of selecting a terrier is;

P(terrier) = 5/11

The probability of selecting a retriever is;

P(retriever) = 2/11

The required probability is

P = 5/11 * 2/11

Evaluate

P = 10/121

Hence, the probability that Rebecka selects a terrier and Aaron selects a retriever is 10/121

Read more about probability at:

brainly.com/question/11234923

#SPJ1

8 0

1 year ago

A student has $78 in his checking account. If the student spends $12 on lunch, which number could be combined with the amount le

Alborosie

Answer:

-66

Step-by-step explanation:

There will be $66 left after the $12 spent on lunch. 66 + -66 = 0

7 0

3 years ago

Read 2 more answers

Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A

iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

$\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA$

7 0

3 years ago

Is (1, 10) a solution to this system of inequalities?<br> y > 2x + 3<br> y > x + 2

Nat2105 [25]

Answer:

yes

Step-by-step explanation:

5 0

2 years ago