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3 years ago

14

One hose can fill a small swimming pool in 95 minutes. A larger hose can fill the pool in 30 minutes. How long will it take the

two hoses to fill the pool working
together?

2 answers:

lianna [129]3 years ago

8 0

Answer:

45 mins for two small hoses and 15 mins for two large ones

Step-by-step explanation:

Lemur [1.5K]3 years ago

3 0

Answer:

I believe this is 65 minutes. if you subtract 30 minutes from 95 you have 65

Step-by-step explanation:

You might be interested in

Which expression has a positive quotient?

svp [43]

Answer:

Only 1st option Negative three-fourths divided by Negative two-thirds has positive quotient.

Option A is correct.

Step-by-step explanation:

We will solve to find out Which expression has a positive quotient.

We know the division rule: $\frac{-a}{-b}=\frac{a}{b}$ and $\frac{-a}{b}=\frac{-a}{b}$

We will use these rules

1) Negative three-fourths divided by Negative two-thirds

$=\frac{-\frac{3}{4} }{-\frac{2}{3} }\\=-\frac{3}{4}\times -\frac{3}{2}\\=\frac{9}{8}$

The quotient is positive.

2) Negative StartFraction 1 over 8 EndFraction divided by 3 and one-fifth

$\frac{-\frac{1}{8} }{3\frac{1}{5} }\\=\frac{-\frac{1}{8} }{\frac{16}{5} }\\=-\frac{1}{18} \times \frac{5}{16}\\=-\frac{5}{288}$

The quotient is negative.

3) 2 and StartFraction 2 over 7 EndFraction divided by negative one-fifth

$\frac{2\frac{2}{7} }{-\frac{1}{5} }\\= \frac{\frac{16}{7} }{-\frac{1}{5} }\\=\frac{16}{7} \times -5\\=-\frac{80}{7}$

The quotient is negative.

4) Negative 6 divided by Five-thirds

$\frac{-6}{\frac{5}{3} }\\=-6 \times \frac{3}{5}\\=-\frac{18}{5}$

The quotient is negative.

So, only 1st option Negative three-fourths divided by Negative two-thirds has positive quotient.

Option A is correct.

4 0

3 years ago

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The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

6 0

2 years ago

Read 2 more answers

How do you do this we need some really big help so help us please

rewona [7]

Well that's division.
•36 divided by 9 is equal to 4.
• 81 divided by 9 is equal to 9.
•45 divided by 5 is equal to 9.
•56 divided by 7 is equal to 8.
•42 divided by 7 is equal to 6.

7 0

3 years ago

Solve the equation. y + 6 = –3y + 26

stiks02 [169]

Answer:

Step-by-step explanation:

3y+y = 26-6

4y = 20

4y/4 = 20/4

y = 5

4 0

3 years ago

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How do you solve substitution to solve the sytem <br> x+y=2<br> -2x+y=2

lord [1]

So you want to make one of the equations equal to one of the placeholders exg y=x+3

so we have
x+y=2 and
-2x+y=2

I chose to make -2x+y=2 equal to y so
-2x+y=2
add 2x to both sides
y=2+2x

subsitute 2+2x for y in x+y=2
x+2+2x=2
add like terms
3x+2=2
subtract 2 from both sides
3x=0
x=0

subsitute x=0 in x+y=2
0+y=2
y=2

x=0
y=2

5 0

3 years ago

Read 2 more answers