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4 years ago

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Help with geometry!!!!!!!!!!!!

1 answer:

slamgirl [31]4 years ago

7 0

Answer:

ok what is it

Explanation:

hi i am Elena Sweet but yu can call me Elena

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AAS theorem<br> SSS postulate <br> SAS postulate<br> ASA postulate<br> Which one?

valentina_108 [34]

Answer:

ASA postulate

Step-by-step explanation:

You have 2 angles that are congruent and also 1 side is congruent

BUT

the side(S) must be in between the angles

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3 years ago

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The measure of an angle is 52 more than its complement. what is the measure of the angle

STALIN [3.7K]

Complement = 90
90-52= 38

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3 years ago

How to find the diameter of a circle with pi

kifflom [539]

Step-by-step explanation: Divide the circumference (perimeter) by pi. That gives you the diameter. Half of that is the radius. To get the area, square the radius, and then multiply by pi.

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3 years ago

Sum to n terms of each of following series. (a) 1 - 7a + 13a ^ 2 - 19a ^ 3+...

julia-pushkina [17]

Notice that the difference in the absolute values of consecutive coefficients is constant:

|-7| - 1 = 6

13 - |-7| = 6

|-19| - 13 = 6

and so on. This means the coefficients in the given series

$\displaystyle \sum_{i=1}^\infty c_i a^{i-1} = \sum_{i=1}^\infty |c_i| (-a)^{i-1} = 1 - 7a + 13a^2 - 19a^3 + \cdots$

occur in arithmetic progression; in particular, we have first value $c_1 = 1$ and for $n>1$ , $|c_i|=|c_{i-1}|+6$ . Solving this recurrence, we end up with

$|c_i| = |c_1| + 6(i-1) \implies |c_i| = 6i - 5$

So, the sum to $n$ terms of this series is

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \underbrace{\sum_{i=1}^n i (-a)^{i-1}}_{S'} - 5 \underbrace{\sum_{i=1}^n (-a)^{i-1}}_S$

The second sum $S$ is a standard geometric series, which is easy to compute:

$S = 1 - a + a^2 - a^3 + \cdots + (-a)^{n-1}$

Multiply both sides by $-a$ :

$-aS = -a + a^2 - a^3 + a^4 - \cdots + (-a)^n$

Subtract this from $S$ to eliminate the intermediate terms to end up with

$S - (-aS) = 1 - (-a)^n \implies (1-(-a)) S = 1 - (-a)^n \implies S = \dfrac{1 - (-a)^n}{1 + a}$

The first sum $S'$ can be handled with simple algebraic manipulation.

$S' = \displaystyle \sum_{i=1}^n i (-a)^{i-1}$

$\displaystyle S' = \sum_{i=0}^{n-1} (i+1) (-a)^i$

$\displaystyle S' = \sum_{i=0}^{n-1} i (-a)^i + \sum_{i=0}^{n-1} (-a)^i$

$\displaystyle S' = \sum_{i=1}^{n-1} i (-a)^i + \sum_{i=1}^n (-a)^{i-1}$

$\displaystyle S' = \sum_{i=1}^n i (-a)^i - n (-a)^n + S$

$\displaystyle S' = -a \sum_{i=1}^n i (-a)^{i-1} - n (-a)^n + S$

$\displaystyle S' = -a S' - n (-a)^n + \dfrac{1 - (-a)^n}{1 + a}$

$\displaystyle (1 + a) S' = \dfrac{1 - (-a)^n - n (1 + a) (-a)^n}{1 + a}$

$\displaystyle S' = \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2}$

Putting everything together, we have

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 S' - 5 S$

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2} - 5 \dfrac{1 - (-a)^n}{1 + a}$

$\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} =\boxed{\dfrac{1 - 5a - (6n+1) (-a)^n + (6n-5) (-a)^{n+1}}{(1+a)^2}}$

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2 years ago

Which function represents the graph below?

puteri [66]

Answer:

b

Step-by-step explanation:

Since it intercepts y at 4, and x at 2, we use y=ax+b and rearrange to get the equation from y=2x+4 to 2x-y=-4

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3 years ago

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