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Tomtit [17]
3 years ago
5

In the past year Maya watched 34 movies that she thought were very good. She watched 40 movies over the whole year. Of the movie

s she watched, what percentage did she rate as very good?
Mathematics
1 answer:
kvasek [131]3 years ago
6 0

Answer:

90 percent

Step-by-step explanation:

the fraction is 90/100 so the percentage is 90%

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Given that lines m and n are parallel and m<60, find the measures of the remaining angles m<3=?
grigory [225]

When you have two parallel lines being cut by a transversal, you can use the next statements for the angles:

The measurement of the angles in green is the same (are congruents)

The measurement of the angles in blue is the same (are congruents)

An angle in green is a complementary angle with an angle in blue. (The addition of the two angles has to be 180º)

You know that:

m\angle2=60º

As the angle 3 is congruent with angle 2:

m\angle3=60º

6 0
1 year ago
What is the graph of the absolute value equation? Y=-2|x|+5
dlinn [17]
X:-2     y:1
x:-1     y:3
x:0      y:5
x:1      y:3
x:2      y:1

 I hope this helps you :)
3 0
3 years ago
What is the best price?
skad [1K]

Answer:

1 bag

Step-by-step explanation:

because the other ones are more expensive

3 0
3 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
The value x = 3 is a solution to each of the following except which?
ololo11 [35]

Answer:

x \: is \: not \: a \: solution \: in \to \: 5x + 4 < 19. \\

Step-by-step explanation:

\: if \: x = 3 \: then \\  4x-1=2x +5  \: is \: same \: as  \to \:  \\ 4(3) - 1  = 2(3) + 5\\ 12 - 1 = 6 + 5 \\ 11 =11 \to \: \\  \boxed{ x \: is \:a \:  solution \: here}  \\  \\ if \: x = 3 \\ then \to \\ 5x + 4    <   19\: is \: same \:as \to \\ 5(3) + 4 < 19 \\ 15 + 4 < 19 \\ 19 \: is \: equal \: to \: 19 : \\ 19 \:  can \: not \: be \: less \: than \: 19 \\ \boxed{ henc e\: x \: is \: not \: a \: solution \: here}

♨Rage♨

6 0
3 years ago
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