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Arada [10]
3 years ago
12

Find Slope (3,2),(-6,2)

Mathematics
2 answers:
NikAS [45]3 years ago
8 0

By using slope formula (x2-x1) ^2 / (y2-y1) ^2 i.e

x1 = 3          x2 = -6

y1 = 2          y2 =  2

Using slope formula,

Slope = {(-6) - 3} ^2 / (2 - 2 ) ^2

*use this and solve and if this was helpful let me know.

         

dlinn [17]3 years ago
6 0

slope=0

it's a straight horizontal line when you graph it.

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What will be the value of
madreJ [45]

The expression as given doesn't make much sense. I think you're trying to describe an infinitely nested radical. We can express this recursively by

\begin{cases}a_1=\sqrt{42}\\a_n=\sqrt{42+a_{n-1}}\end{cases}

Then you want to know the value of

\displaystyle\lim_{n\to\infty}a_n

if it exists.

To show the limit exists and that a_n converges to some limit, we can try showing that the sequence is bounded and monotonic.

Boundedness: It's true that a_1=\sqrt{42}\le\sqrt{49}=7. Suppose a_k\le 7. Then a_{k+1}=\sqrt{42+a_k}\le\sqrt{42+7}=7. So by induction, a_n is bounded above by 7 for all n.

Monontonicity: We have a_1=\sqrt{42} and a_2=\sqrt{42+\sqrt{42}}. It should be quite clear that a_2>a_1. Suppose a_k>a_{k-1}. Then a_{k+1}=\sqrt{42+a_k}>\sqrt{42+a_{k-1}}=a_k. So by induction, a_n is monotonically increasing.

Then because a_n is bounded above and strictly increasing, the limit exists. Call it L. Now,

\displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n-1}=L

\displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty}\sqrt{42+a_{n-1}}=\sqrt{42+\lim_{n\to\infty}a_{n-1}}

\implies L=\sqrt{42+L}

Solve for L:

L^2=42+L\implies L^2-L-42=(L-7)(L+6)=0\implies L=7

We omit L=-6 because our analysis above showed that L must be positive.

So the value of the infinitely nested radical is 7.

4 0
3 years ago
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Anna [14]

Answer:

18.

Step-by-step explanation:

3 x 6 = 18. He eats 3 slices every 5 minutes so 5 minutes x 6 is half an hour. so 18 is your answer.

6 0
3 years ago
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Answer:

(x,y) maps to (-y,x).

Step-by-step explanation:

Use vertex I as an example.

Pre-image I is graphed at point (-3,2), while image I' is graphed at point (-2,-3) .

In order to change between these points, the x and y coordinates must be swapped, as well as multiplying the y coordinate by -1.

8 0
3 years ago
Could someone please help and explain how to do this
LiRa [457]

Answer:

Step-by-step explanation:

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Neporo4naja [7]

Answer:

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Step-by-step explanation:

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6 0
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