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2 years ago

Name at least 3 angles for each diagram

Mathematics

2 answers:

irga5000 [103]2 years ago

8 0

Answer:

angle RSU.

angle RST

angle UST

seraphim [82]2 years ago

4 0

Answer:

Acute angle, Obtuse angle, and Zero angle

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Hey! Giving away 50 free points! Just type "octopus" as an answer and I will give you 5 star, a heart, and brainly will give you

Arturiano [62]

Answer:

octopus

Step-by-step explanation:

thank you so muchhhhh

3 0

3 years ago

Read 2 more answers

Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

7 0

2 years ago

Please help with this, i tried to solve for x, and it didnt give me any of those answers

Firdavs [7]

I hope this helps you

4.1+4. (-x)+2x=3.x+3.1

4-4x+2x=3x+3

-2x-3x=3-4

-5x= -1

x=1/5

3 0

2 years ago

Brenda took out a personal loan for $12,000 at an interest rate of 12% compounded monthly. She made arrangements to pay the loan

iragen [17]

Answer: her monthly payments would be $267

Step-by-step explanation:

We would apply the periodic interest rate formula which is expressed as

P = a/[{(1+r)^n]-1}/{r(1+r)^n}]

Where

P represents the monthly payments.

a represents the amount of the loan

r represents the annual rate.

n represents number of monthly payments. Therefore

a = $12000

r = 0.12/12 = 0.01

n = 12 × 5 = 60

Therefore,

P = 12000/[{(1+0.01)^60]-1}/{0.01(1+0.01)^60}]

12000/[{(1.01)^60]-1}/{0.01(1.01)^60}]

P = 12000/{1.817 -1}/[0.01(1.817)]

P = 12000/(0.817/0.01817)

P = 12000/44.96

P = $267

7 0

2 years ago

Solve this emergency i will mark you as brainliest

Umnica [9.8K]

Answer:

3/8 cups.

Step-by-step explanation:

2 tablespoons of sugar = 1/16 * 2 = 1/8 cup of sugar.

1/2 cup = 4/8 cups.

So the extra sugar required = 4/8 - 1/8

= 3/8 cups.

3 0

3 years ago

Read 2 more answers