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3 years ago

Q4.

Mathematics

2 answers:

xenn [34]3 years ago

3 0

Answer:

Step-by-step explanation:

loris [4]3 years ago

3 0

Answer:

£2478.37 this is what I got, I'm not too sure whether it is right though

Step-by-step explanation:

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Hannah has another candle that is 14 cm tall. How fast must it burn in order to also be 6 cm tall after 4 hours? Explain your th

RideAnS [48]

Answer:

The candle must burn at 2cm per hour in order to be 6cm tall after 4 hours. I know this because the candle has to burn 8cm in 4 hours and that amounts to 2cm per hour.

Step-by-step explanation:

Lets solve!

We have to find out how much of the candle needs to be burned!

14cm - 6cm = 8cm

The candle must burn 8cm in 4 hours!

Lets find out how much the candle burns in 1 hour.

$\frac{8cm}{4hours}$ = 2cm per hour

The candle must burn at 2cm per hour in order to be 6cm tall after 4 hours. I know this because the candle has to burn 8cm in 4 hours and that amounts to 2cm per hour.

Woohoo, We did it! <u>Would you like to mark my answer as brainliest?</u> I would love that!

6 0

3 years ago

Write the sentence as an equation.<br> 143 more than the product of g and 369 is equal to 141.

DiKsa [7]

Answer: 143>(369g=141)

Step-by-step explanation:

143>(369g=141)

3 0

3 years ago

How many different books did the children read

Alex777 [14]

Answer:

16 books

Step-by-step explanation:

if Sandra read 5 books and Deacon read 6 and Breanna read 7 and they all one same book, then that means sandra read 4 different books then Deacon and Breanna. Deacon read 5 different books then Sandra and Breanna. Breanna read 6 different books then Deacon and Sandra.

So Sandras 4 books plus Deacons 5 books plus Breannas 6 books equals up to 15 but, now you need to add the one book they all read which makes the answer 16.

(Hopes this helps)

3 0

3 years ago

A plumber charges an initial fee of $60 and $15 per hour. Write an equation that represents how much the plumber charges, y, if

34kurt

Answer:

Y=15x+60

Step-by-step explanation:

6 0

2 years ago

Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ

atroni [7]

Answer: $y=Ce^(^3^t^{^9}^)$

Step-by-step explanation:

Beginning with the first differential equation:

$\frac{dy}{dt} =27t^8y$

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

$\frac{1}{y} \frac{dy}{dt} =27t^8$

Multiply both sides by 'dt' to get:

$\frac{1}{y}dy =27t^8dt$

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

$\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt$

$ln(y)=27(\frac{1}{9} t^9)+C$

$ln(y)=3t^9+C$

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

$e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)$

$y=e^(^3^t^{^9} ^+^C^)$

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

$y=e^(^3^t^{^9}^)e^C$

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

$y=Ce^(^3^t^{^9}^)$

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

$\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8$

Now check if the derivative equals the right side of the original differential equation:

$(Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)$

$Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)$

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.

7 0

2 years ago