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3 years ago

How do you solve w=120(1/3) by step

Mathematics

1 answer:

lubasha [3.4K]3 years ago

6 0

Answer:

w=40

Step-by-step explanation:

120*1/3=40

Therefore, w=40

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The mid-point of the straight line joining A(4,-3), B(6,-2) is located at C.

baherus [9]

I hope this helps you

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3 years ago

Marietta was given a raise of $0.75 per hour, which gave her a new wage of $12.25 per hour. Write and solve an equation to deter

guajiro [1.7K]

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4 years ago

Read 2 more answers

1. What is the slope and how is it determined from a g raph?

Goryan [66]

Answer:

The<u> rise divided by the run </u>is the slope of a line. The rise and run are used to estimate the slope of a line from its graph.

To determine the intercepts from a graph, We <u>put y equal to zero</u> and solve for x to find the x-intercept. Similarly, we put x equal to zero and solve for y to find the y-intercept.

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3 years ago

Consider the following numbers: 7000, 700, 70, 0.7, 0.07, 0.007

masya89 [10]

Answer:

7000= 7 x 10^{3}, 700= 7 x 10^{2}, 70= 7 x 10^{1}, 0.7= 7 x 10^{-1}, 0.07= 7 x 10^{-2}, 0.007= 7 x 10^{-3}

Step-by-step explanation:

YES

6 0

4 years 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

jenyasd209 [6]

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

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

3 0

3 years ago