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Ugo [173]
3 years ago
5

Please help this is mainly on fuctions​

Mathematics
1 answer:
frozen [14]3 years ago
8 0

Answer:

B and C

Step-by-step explanation:

We are going to look at each choice:

Let's look at choice A.

f(5)=1

f(5) means look at function f and then see which piece contains x=5.

5>1 so we are looking at x^2 for x=5 which gives you 5^2=25, this is not 1.

The answer is not A.

Let's look at choice B.

f(2)=4

f(2) means look at function f and then see which piece contains x=2.

2>1 so we are looking at x^2 for x=2 which gives you 2^2=4, so this is true that f(2)=4.

Let's look at choice C.

f(1)=5

f(1) means look at function f and then see which piece contains x=1.

1=1 which means we are using 5 for x=1, so f(1)=5 which is what choice C says so choice C is true.

Let's look at choice D.

f(-2)=4

f(-2) means look at function f and then see which piece contains x=-2

-2<1 so we use 2x for x=-2, so we get 2(-2)=-4 which is not 4 so f(-2)=4 is not true.

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5 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
If Sam buys 200 stickers at $10 and sells them for $.25 each how much is he making?
Alex17521 [72]
First you will need to multiple 200 by .25 to see how much money Sam will make.
200 x .25 = $50 
now you need to subtract that price by 10 because he spent $10 to get the stickers.
50 - 10 = $40
Sam is making $40 off the stickers.
let me know if you have any other questions
:)
3 0
3 years ago
What’s the answer I don’t get this
luda_lava [24]

Answer:

K(-1)=6

Step-by-step explanation:

t=-1

2 times -1 squared

-1 squared=1

2x1 = 2

2+4=6

K(-1)=6

5 0
3 years ago
The critical points of a rational inequality are x = –4 and x = 2. Which set of points can be tested to find a complete solution
katovenus [111]

Answer:

-2

Step-by-step explanation:

3 0
3 years ago
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