What is 2 divided 1 6/7 ? Please answer

1 answer:

6 0

If you would like to divide 2 by 1 6/7, you can do this using the following steps:

1 6/7 = 13/7

2 / 1 6/7 = 2 / 13/7 = 2/1 / 13/7 = 2/1 * 7/13 = 14/13 = 1 1/13

The correct result would be 1 1/13.

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Today is the Harperville Fall for Fun festival, and Naomi is volunteering at the balloon stand. When she practiced last night, N

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Answer:

Step-by-step explanation:

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1 year ago

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

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Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

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4C. Quintin is using the three different shaped

Sauron [17]

The smallest number of tiles Quintin will need in order to tile his floor is 20

The given parameters;

number of different shapes of tiles available = 3

number of each shape = 5

area of each square shape tiles, A = 2000 cm²

length of the floor, L = 10 m = 1000 cm

width of the floor, W = 6 m = 600 cm

To find:

the smallest number of tiles Quintin will need in order to tile his floor

Among the three different shapes available, total area of one is calculated as;

$A_{one \ square \ type} = 5 \times 2000 \ cm^2 = 10,000 \ cm^2$

Area of the floor is calculated as;

$A_{floor} = 1000 \ cm \times 600 \ cm = 600,000 \ cm^2$

The maximum number tiles needed (this will be possible if only one shape type is used)

$maximum \ number= \frac{Area \ of \ floor}{total \ area \ of \ one \ shape \ type} \\\\maximum \ number= \frac{600,000 \ cm^2}{10,000 \ cm^2} \\\\maximum \ number= 60$