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Yanka [14]
3 years ago
11

HELP ME PLEASE!!!

Mathematics
1 answer:
Pavel [41]3 years ago
7 0

Answer:

Lo siento amigo mío pero ha llegado el momento. Necesitas hacer tu propia tarea porque la profunda y oscura verdad es ... nadie te ayudará cuando falles en la prueba. Por eso estoy aquí para llevarte. Deja de pedir respuestas y pruébalas tú mismo. Pídele ayuda a tu maestro. Pero deja de pedirle respuestas a la gente. Eso es definitivamente una trampa.

Step-by-step explanation:

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Y-intercepts and X-intercepts
MAVERICK [17]

Answer:

the bottom with negative numbers are the point where you will start from, so for example if the problem in -4,1/2 then the starting point will be at -4 and the half will be on the middle line. if you don't understand and you have a more specific question reply with it!

Step-by-step explanation:

4 0
3 years ago
If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.
Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have S_1=3\times2^{1-1}+2(-1)^1=3-2=1, which agrees with the initial value <em>S</em>₁ = 1.

• Assume the closed form is correct for all <em>n</em> up to <em>n</em> = <em>k</em>. In particular, we assume

S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}

and

S_k=3\times2^{k-1}+2(-1)^k

We want to then use this assumption to show the closed form is correct for <em>n</em> = <em>k</em> + 1, or

S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}

From the given recurrence, we know

S_{k+1}=S_k+2S_{k-1}

so that

S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)

S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}

S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)

S_{k+1}=3\times2^k-2(-1)^k

S_{k+1}=3\times2^k+2(-1)(-1)^k

\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}

which is what we needed. QED

6 0
3 years ago
Can you help me i don’t know the answers
Advocard [28]
1)

An irrational number is a number that a) can't be written as a fraction of two whole numbers AND b) is an infinite decimal without any sort of pattern.

For the first answer choice, clearly \frac{1}{3} does not pass the first criterion so we look at the second choice.

Let's come back to \sqrt{2} and \pi.

\frac{2}{9} doesn't meet our first criterion, and let's skip \sqrt{3} for now.

It is often easier to disprove an irrational number than to prove one. There are a few famous irrationals to know (although there is an infinite number of irrationals). The most common are \sqrt{2},  \pi, e,  \sqrt{3}. For now, it's just helpful to know these and recognize them.

So we can check off \sqrt{2},  \pi and \sqrt{3}.

2) 

For this next question, we know that \sqrt{64} = 8. Clearly this isn't irrational. Likewise, \frac{1}{2} isn't irrational. \frac{16}{4} =  \frac{4}{4} = 1, which is rational, leaving only \frac{ \sqrt{20}}{5} =  \frac{2 \sqrt{5} }{5}. By process of elimination, this is the correct answer. Indeed, \sqrt{5} is an irrational number.

3) This notation means that we have 0.3636363636... and so on, to an infinite number of digits. It is called a repeating decimal.

But it can be written as a fraction because its pattern repeats, unlike for an irrational number.

Let's say x=0.36363636.... Would you agree that 100x=36.36363636...? (We choose to multiply by 100 because there are two decimals that repeat. For 1, choose 10, for 3 choose 1,000, and so on.)

Now, let's subtract x from 100x and solve.

100x=36.36363636\\-x \ \ \ \ \ \ \ -0.36363636\\99x=36\\\\x= \dfrac{36}{99}= \dfrac{4}{11}

Voila!
4 0
3 years ago
Read 2 more answers
Which container holds more, a Elgin millimeter carton, or a 1 L bottle
anyanavicka [17]
The answer would be a 1 Liter bottle, because a liter is 33 oz rounded and and Elgin millimeter container is 25 oz.
7 0
3 years ago
What does inverse operations mean?
jok3333 [9.3K]

Answer:

A pair of inverse operations is defined as two operations that will be performed on a number or. variable, that always results in the original number or variable. Another way to think of this is. that the two inverse operations “undo” each other.

Step-by-step explanation:

for example addition and subtraction are and divison and multiplication are, because they inverse each other.

8 0
2 years ago
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