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3 years ago

At the bank of a river, by measurement, a = 18 ft. b = 30 ft. c = 24 ft. d = 14 ft. How long is x?

2 answers:

5 0

Since you did not give the picture of the problems so it is difficult to solve the problem. but i think this problem is about similar triangles, similar triangle are triangles that have corresponding angles are congruent and their sides are in proportion.
( x / 24) = 30 /18
x = 24 ( 30 / 18 )
x = 40 ft

hram777 [196]3 years ago

5 0

This might get taken down but I just want to point out how impressive it is that the person above this solved it without the picture

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What is the circumference of a 12-inch diameter pizza? Leave the answer in terms of ?.

adoni [48]

Answer:

C = 37.7 inches

Step-by-step explanation:

You are looking for the circumference.

Formula for the circumference-> C=2πr

variables:

c = circumference

r = radius

You have the diameter, which is 12 inches. Radius is half of the diameter. which means r is half of 12 (aka 6).

put the radius where the variable is in the formula.

C=2π(6)

you now have enough information to solve for circumference.

C = 2 * π * 6

C = 37.7

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3 years ago

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Alenkinab [10]

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2 years ago

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

Three screenshots, answer em all pls

Shtirlitz [24]

I just did the second one, the one that looks this a side ways L,
here are the points
(4,4). (5,4). (3,1). (3,2). (4,2). hope this helped

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2 years ago

1/3(15y-6) I need help plz!!!

Rufina [12.5K]

Answer:

(4.99y-1.99); rounded it would be (5y-2)

Step-by-step explanation:

Distribute 1/3 into the parentheses, since you can't subtract 15y-6 because they don't have the same variable.

5 0

2 years ago