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3 years ago

11

A rectangular fish tank has a base that is 8 inches by 7 inches. How much water will it take to fill the tank to a depth of 5 in

ches?

1 answer:

Blizzard [7]3 years ago

5 0

V=LWH
L=8
W=7
H=5

V=8*7*5=280 in^3

you need 280 in^3 of water

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*^%Brainliest 15 pounds Question 5 (2 points)

galben [10]

Answer:

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Step-by-step explanation:

Separate into two parts

$\frac{4xy^{3} }{2xy^{2} } + \frac{8x^{2} {y}^{5} }{2x {y}^{2} }$

Simplify:

This is for the first fraction

$\frac{4x {y}^{3} }{2x {y}^{2} } = (4 \div 2)(x \div x)( {y}^{3} \div {y}^{2} )$

$2(1)( {y}^{3 - 2} )$

$2{y}$

Now for the second fraction:

$\frac{8 {x}^{2} {y}^{3} }{2x {y}^{2} } = ( 8 \div 2)( {x}^{2} \div x)( {y}^{5} \div {y}^{2} )$

$4( {x}^{2 - 1} )( {y}^{5 - 2} )$

$4x {y}^{3}$

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$2y + 4x {y}^{3}$

To turn this into the said formula, that would become:

$2 {x}^{0} {y}^{1} + 4 {x}^{1} {y}^{3}$

Where:

a=0

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c=1

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2 years ago

A bacteria culture starts with 12,000 bacteria and the number doubles every 50 minutes.

Vlada [557]

Answer:

a) $y=12000(2)^{\frac{t}{50}}$

b) Approx. 27,569 bacteria

c) About 103 minutes

Step-by-step explanation:

a)

This will follow exponential modelling with form of equation shown below:

$y=Ab^{\frac{t}{n}}$

Where

A is the initial amount (here, 12000)

b is the growth factor (double, so growth factor is "2")

n is the number of minutes in which it doubles, so n = 50

Substituting, we get our formula:

$y=Ab^{\frac{t}{n}}\\y=12000(2)^{\frac{t}{50}}$

b)

To get number of bacteria after 1 hour, we have to plug in the time into "t" of the formula we wrote earlier.

Remember, t is in minutes, so

1 hour = 60 minutes

t = 60

Substituting, we get:

$y=12000(2)^{\frac{t}{50}}\\y=12000(2)^{\frac{60}{50}}\\y=12000(2)^{\frac{6}{5}}\\y=27,568.76$

The number of bacteria after 1 hour would approximate be <u>27,569 bacteria</u>

<u></u>

c)

To get TIME to go to 50,000 bacteria, we will substitute 50,000 into "y" of the equation and solve the equation using natural logarithms to get t. Shown below:

$y=12000(2)^{\frac{t}{50}}\\50,000=12,000(2)^{\frac{t}{50}}\\4.17=2^{\frac{t}{50}}\\Ln(4.17)=Ln(2^{\frac{t}{50}})\\Ln(4.17)=\frac{t}{50}*Ln(2)\\\frac{t}{50}=\frac{Ln(4.17)}{Ln(2)}\\\frac{t}{50}=2.06\\t=103$

After about 103 minutes, there will be 50,000 bacteria

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