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Pachacha [2.7K]

2 years ago

11

GIVING BRAINLIEST!!!!!

2 answers:

Jet001 [13]2 years ago

7 0

plane PBC is right answer.

it's a logical answer....

yaroslaw [1]2 years ago

5 0

Answer:

Plane PBC

Step-by-step explanation:

Plane PBC contains 4 points : P, B, C, and Q.

The question asks which contains all three of C, Q, and P.

The plane is Plane PBC

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Hi please help i’ll give brainliest

svetlana [45]

Answer:

It's winter

Step-by-step explanation:

because if the south pole is oriented towards the sun it's summer what's the oopsite of summer it's winter

6 0

3 years ago

Which equation represents a line that has a slope of Negative one-half and passes through point (4, –5)? y = negative one-half x

Anni [7]

Answer:

y = (-1/2)x - 3

Step-by-step explanation:

Please present the equation symbolically instead of in words:

m = -1/2, passing through (4, -5):

Using the slope-intercept formula y = mx + b, and using m = -1/2, x = 4 and y = -5, we get:

-5 = (-1/2)(4) + b, or -5 = -2 + b. Then b = -3 and the desired equation is:

y = (-1/2)x - 3

3 0

3 years ago

Read 2 more answers

The points (5,r) and (3,−2) lie on a line with a slope of −2 . What is the value of r ?

Katen [24]

Answer:

<u>r = -</u><u>3</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u>......

8 0

3 years ago

Read 2 more answers

Https://brainly.com/question/21130598<br> pLS ANSWER

Sladkaya [172]

There is no question. I’m sorry :(. Can you mark me as brainliest so I can help more people !!?

6 0

3 years ago

The sum of the first n terms of a geometric series is 364? The sum of their reciprocals 364/243. If the first term is 1, find n

Afina-wow [57]

If the geometric series has first term $a$ and common ratio $r$ , then its $N$ -th partial sum is

$\displaystyle S_N = \sum_{n=1}^N ar^{n-1} = a + ar + ar^2 + \cdots + ar^{N-1}$

Multiply both sides by $r$ , then subtract $rS_N$ from $S_N$ to eliminate all the middle terms and solve for $S_N$ :

$rS_N = ar + ar^2 + ar^3 + \cdots + ar^N$

$\implies (1 - r) S_N = a - ar^N$

$\implies S_N = \dfrac{a(1-r^N)}{1-r}$

The $N$ -th partial sum for the series of reciprocal terms (denoted by $S'_N$ ) can be computed similarly:

$\displaystyle S'_N = \sum_{n=1}^N \frac1{ar^{N-1}} = \frac1a + \frac1{ar} + \frac1{ar^2} + \cdots + \frac1{ar^{N-1}}$

$\dfrac{S'_N}r = \dfrac1{ar} + \dfrac1{ar^2} + \dfrac1{ar^3} + \cdots + \dfrac1{ar^N}$

$\implies \left(1 - \dfrac1r\right) S'_N = \dfrac1a - \dfrac1{ar^N}$

$\implies S'_N = \dfrac{1 - \frac1{r^N}}{a\left(1 - \frac1r\right)} = \dfrac{r^N - 1}{a(r^N - r^{N-1})} = \dfrac{1 - r^N}{a r^{N-1} (1 - r)}$

We're given that $a=1$ , and the sum of the first $n$ terms of the series is

$S_n = \dfrac{1-r^n}{1-r} = 364$

and the sum of their reciprocals is

$S'_n = \dfrac{1 - r^n}{r^{n-1}(1 - r)} = \dfrac{364}{243}$

By substitution,

$\dfrac{1 - r^n}{r^{n-1}(1-r)} = \dfrac{364}{r^{n-1}} = \dfrac{364}{243} \implies r^{n-1} = 243$

Manipulating the $S_n$ equation gives

$\dfrac{1 - r^n}{1-r} = 364 \implies r (364 - r^{n-1}) = 363$

so that substituting again yields

$r (364 - 243) = 363 \implies 121r = 363 \implies \boxed{r=3}$

and it follows that

$r^{n-1} = 243 \implies 3^{n-1} = 3^5 \implies n-1 = 5 \implies \boxed{n=6}$

5 0

2 years ago