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3 years ago

Which expression is equivalent to 3V64k^12

Mathematics

1 answer:

scoundrel [369]3 years ago

4 0

Answer:

I think the answer is C

Step-by-step explanation:

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If the length of rectangle is thrice of its breadth and it's perimeter is 32 cm then finds its area.

jeka57 [31]

Answer:

48 cm²

Explanation:

Let the breadth be b, then the length is 3b

P = 2(Length + Breadth)

32 = 2(3b + b)

32 = 2(4b)

8b = 32

b = 4

Breadth is 4 cm

Length: 3b = 3(4) = 12 cm

Area of rectangle:

Length × Breadth

12 × 4

48 cm²

3 0

2 years ago

Read 2 more answers

Solve for y. –0.8y + 5.49 = –19.86 − 2.3y y =

anastassius [24]

Answer:

y = -16.9

Step-by-step explanation:

→Add 0.8y to both sides:

-0.8y + 5.49 = -19.86 - 2.3y

5.49 = -19.86 - 1.5y

→Add 19.86 to both sides:

25.35 = -1.5y

→Divide both sides by -1.5:

-16.9 = y

8 0

3 years ago

Read 2 more answers

A Turtle Crawled 0.06617 kilometers. How many millimeters is this?

Paul [167]

Answer:

66170 mm

Step-by-step explanation:

0.06617 km * 1000 km/m * 1000 m/mm = 66170 mm

4 0

3 years ago

Is a three-digt whole number

Rudiy27

Answer:

693

Step-by-step explanation:

7*9*11*3=693

5 0

3 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