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2 years ago

These two lines represent a system of equations. What is the solution to that system? Help ?

Mathematics

2 answers:

Mars2501 [29]2 years ago

8 0

Answer:

yhuvguvcxdcfvgbhnjmkj

Step-by-step explanation:

blsea [12.9K]2 years ago

3 0

Answer: (-1,1)

Step-by-step explanation: Look at the intersection point of the lines

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What is the solution to the system of equation y=2/3x+3 x=-2

Leya [2.2K]

Answer

Is x equal to negative 2?

Step-by-step explanation:

7 0

3 years ago

Find each of the following and explain your reasoning: 1) The length of segment BE

777dan777 [17]

Answer:

BE = 2.5

Step-by-step explanation:

find the midpoint of AD

$(\frac{0+4}{2},\frac{2+7}{2})\\(2,4.5)\\BE=2.5$

4 0

3 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

Mr. mole left his burrow

pav-90 [236]

Answer:Did he ever come back?

Step-by-step explanation:

8 0

3 years ago

(Score for Question 3: ___ of 5 points)

mote1985 [20]

Here are the step for solving this:
1. 3y + 5 > 23
-5 -5
3y/3 > 18/3
y > 6

In words what you are doing is getting y by itself on one side. Subtract the 5 to undo the adding of 5 first. Then, divide both sides by 3.

7 0

3 years ago