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EastWind [94]
3 years ago
9

HELP PLEASE ASAP!!!!!

Mathematics
2 answers:
Lina20 [59]3 years ago
7 0

Answer:

1 1/2 cups of flour :))

Step-by-step explanation:

liberstina [14]3 years ago
7 0

Answer:

2 5/8 cups ur welcome

you convert  1 3/4 to improper fraction so 7/4. then divide 7/4 by 2 which is 0.875. 0.875+7/4= 2.625 which is 2 2/3.

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Twenty-seven is the product of three and a number
n200080 [17]

It’s the product of three and nine.


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3 years ago
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Plz help me and show work this is worth 25points
andrezito [222]
The answers are..
6. 55
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Use the graph to find the following.
poizon [28]

The values of x at wich F(x) has local minimums are x = -2 and x = 4, and the local minimums are:

  • F(-2) = -3
  • F(4) = -5

<h3>What is a local maximum/minimum?</h3>

A local maximum is a point on the graph of the function, such that in a close vicinity it is the maximum value there. So, on an interval (a, b) a local maximum would be F(c) such that:

c ∈ (a, b)

F(c) ≥ F(x) for ∀ x ∈ [a, b]

A local minimum is kinda the same, but it must meet the condition:

c ∈ (a, b)

F(c) ≤ F(x) for ∀ x ∈ [a, b]

A) We can see two local minimums, we need to identify at which values of x do they happen.

The first local minimum happens at x = -2

The second local minimum happens at x = 4.

B) The local minimums are given by F(-2) and F(4), in this case, the local minimums are:

  • F(-2) = -3
  • F(4) = -5

If you want to learn more about minimums/maximums, you can read:

brainly.com/question/2118500

3 0
2 years ago
Lim (n/3n-1)^(n-1)<br> n<br> →<br> ∞
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, <em>e</em> :

\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/(<em>n</em> - 9) with 1/<em>m</em>, so that

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

From the relation 9<em>m</em> = <em>n</em> - 9, we see that <em>m</em> also approaches infinity as <em>n</em> 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
3 years ago
Describe the transformation that changes triangle ABC to triangle A'B'C
Nonamiya [84]
It is a down shift of 3.
Also written as (x, y-3)
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