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3 years ago

Hlo friends good morning

Mathematics

1 answer:

NikAS [45]3 years ago

8 0

arey. buy deepa notes no need to complicate it

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Bruh do those mean i’ll give you a cookie :>

nirvana33 [79]

Answer:

SAS

And I don’t know if your going to give me a cookie. We need to social distance you know ;)

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3 years ago

Please answer the question. The picture is right below.

zhenek [66]

Hey man I’ve never seen nothing like this but I’d you’re doing math on ap classroom , man , GOOD LUCK

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3 years ago

Point A is at (-7, -9) and point M is at (-0.5, -3).

Anit [1.1K]

Answer:

B ( 6 , 3 )

Step-by-step explanation:

( $x_{1}$ , $y_{1}$ )

( $x_{2}$ , $y_{2}$ )

Coordinates of midpoint are ( $\frac{x_{1} +x_{2} }{2}$ , $\frac{y_{1} +y_{2} }{2}$ )

~~~~~~~~~~~~~~~~~~

A ( - 7 , - 9 )

B ( x , y )

M ( - 0.5 , - 3 )

$\frac{-7+x}{2}$ = - 0.5 ⇒ x = 6

$\frac{-9+y}{2}$ = - 3 ⇒ y = 3

B ( 6 , 3 )

8 0

3 years ago

Please solve the problem with steps

Debora [2.8K]

Answer:

Infinite series equals 4/5

Step-by-step explanation:

Notice that the series can be written as a combination of two geometric series, that can be found independently:

$\frac{3^{n-1}-1}{6^{n-1}} =\frac{3^{n-1}}{6^{n-1}} -\frac{1}{6^{n-1}} =(\frac{1}{2})^{n-1} -\frac{1}{6^{n-1}}$

The first one: $(\frac{1}{2})^{n-1}$ is a geometric sequence of first term ( $a_1$ ) "1" and common ratio (r) " $\frac{1}{2}$ ", so since the common ratio is smaller than one, we can find an answer for the infinite addition of its terms, given by: $Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{2} } =\frac{1}{\frac{1}{2} } =2$

The second one: $\frac{1}{6^{n-1}}$ is a geometric sequence of first term "1", and common ratio (r) " $\frac{1}{6}$ ". Again, since the common ratio is smaller than one, we can find its infinite sum: