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2 years ago

X - y = 4 2x + y = -4 x = y =

Mathematics

1 answer:

Aleks04 [339]2 years ago

8 0

Answer: x = 4 + y

x= -2 - y/2

Step-by-step explanation:add y to both sides of the equation

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n200080 [17]

Answer:

7.9

Step-by-step explanation:

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3 years ago

Instructions: Determine if the two triangles in the image are congruent. If they are, state how you know by identifying the post

Zigmanuir [339]

They are as a result of the side angle side postulate. there is a vertical angle in the center, with a congruent side, and an equal angle (SAS)

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3 years ago

Read 2 more answers

Which of the following is equal to the opposite of −45? −|45| −(−45) −|−45| −45

inessss [21]

Answer:

Step-by-step explanation:

It's B because two negatives equal a positive.

C. The absolute value of -45 is 45, leaving the extra negative to hang on in there ---> -|-45|= 45= (bring the negative back over) -45

D. Self explanatory its -45 alone lol.

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2 years ago

Twice the tens digit of a positive, two-digit integer is 1 less than the ones digit. Reversing the digits increases the number b

Vladimir [108]

It is 16
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4 0

3 years ago

Evaluate the infinite sum:

satela [25.4K]

Consider the k-th partial sum,

$S_k = 1 + \dfrac2\pi + \dfrac3{\pi^2} + \cdots + \dfrac k{\pi^{k-1}}$

More compactly,

$\displaystyle S_k = \sum_{i=1}^k \frac i{\pi^{i-1}} = \frac{(1-\pi)k+\pi^{k+1}-\pi}{(1-\pi)^2\pi^{k-1}}$

(this is just another case of a similar sum you asked about a while ago [24494877])

The infinite sum is the limit of the partial sum as k goes to infinity. We have

$\displaystyle \lim_{k\to\infty} \frac{(1-\pi)k+\pi^{k+1}-\pi}{(1-\pi)^2\pi^{k-1}} = \frac\pi{(1-\pi)^2} \lim_{k\to\infty} \left(\frac{(1-\pi)k}{\pi^k} + \pi - \frac1{\pi^{k-1}} \right) = \boxed{\frac{\pi^2}{(1-\pi)^2}}$

since the non-constant terms in the limit converge to 0.

Alternatively, recall that for |x| < 1, we have

$\dfrac1{1-x} = \displaystyle \sum_{n=0}^\infty x^n$

Differentiating both sides gives

$\dfrac1{(1-x)^2} = \displaystyle \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1}$

also valid for |x| < 1. Take x = 1/π and you get the sum you want to compute.

5 0

2 years ago