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3 years ago

The LCD for the fractions 1/3, 3/4, 5/32, and 8/9 is

Mathematics

1 answer:

Stells [14]3 years ago

5 0

The LCD for the fractions 1/3, 3/4, 5/32, and 8/9 is
<span>B. 288.</span>

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2 years ago

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38.95 * 2 = 77.90.
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At a sale this week, a desk is being sold for $561. This is 68% of the original price. What is the original price?

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Answer:

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Step-by-step explanation:

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2 years ago

Someeee one?????????????????

Ad libitum [116K]

Answer:

Option B) $a_{n} = 2\cdot 4^{n-1}$

Step-by-step explanation:

The given geometric sequence is

2, 8, 32, 128,....

The general form of a geometric sequence is given by

$a_{n} = a_{1}\cdot r^{n-1}$

Where n is the nth term that we want to find out.

a₁ is the first term in the geometric sequence that is 2

r is the common ratio and can found by simply dividing any two consecutive numbers in the sequence,

$r=\frac{8}{2} = 4$

You can try other consecutive numbers too, you will get the same common ratio

$r=\frac{32}{8} = 4$

$r=\frac{128}{32} = 4$

So the common ratio is 4 in this case.

Substitute the value of a₁ and r into the above general equation

$a_{n} = 2\cdot 4^{n-1}$

This is the general form of the given geometric sequence.

Therefore, the correct option is B

Note: Don't multiply the first term and common ratio otherwise you wont get correct results.

Verification:

$a_{n} = 2\cdot 4^{n-1}$

Lets find out the 2nd term

Substitute n = 2

$a_{2} = 2\cdot 4^{2-1} = 2\cdot 4^{1} = 2\cdot 4 = 8$

Lets find out the 3rd term

Substitute n = 3

$a_{3} = 2\cdot 4^{3-1} = 2\cdot 4^{2} = 2\cdot 16 = 32$

Lets find out the 4th term

Substitute n = 4

$a_{4} = 2\cdot 4^{4-1} = 2\cdot 4^{3} = 2\cdot 64 = 128$

Lets find out the 5th term

Substitute n = 5

$a_{5} = 2\cdot 4^{5-1} = 2\cdot 4^{4} = 2\cdot 256 = 512$

Hence, we are getting correct results!

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3 years ago

Use Green's Theorem to calculate the circulation of F =2xyi around the rectangle 0≤x≤8, 0≤y≤3, oriented counterclockwise.

Tamiku [17]

Green's theorem says the circulation of $\vec F$ along the rectangle's border $C$ is equal to the integral of the curl of $\vec F$ over the rectangle's interior $D$ .

Given $\vec F(x,y)=2xy\,\vec\imath$ , its curl is the determinant

$\det\begin{bmatrix}\frac\partial{\partial x}&\frac\partial{\partial y}\\2xy&0\end{bmatrix}=\dfrac{\partial(0)}{\partial x}-\dfrac{\partial(2xy)}{\partial y}=-2x$

So we have

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_D-2x\,\mathrm dx\,\mathrm dy=-2\int_0^3\int_0^8x\,\mathrm dx\,\mathrm dy=\boxed{-192}$

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3 years ago