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2 years ago

Can I get some help please

Mathematics

2 answers:

vagabundo [1.1K]2 years ago

5 0

Answer:

18 - No (18 / 2 = 9)

8 - No (8 / 2 = 4)

20 - No (20 / 2 = 10)

14 - Yes (14 / 2 = 7)

kompoz [17]2 years ago

3 0

Answer is No!................

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Need help with this question bad

andreev551 [17]

Answer:

y = $\frac{4}{3} x$

Step-by-step explanation:

The equation is y = $\frac{4}{3} x$ . To draw the line, take two points (0, 0) and (3, 4). You can draw a straight line connecting those points.

7 0

2 years ago

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A moterboat that ravels with a speen of 20km/hour in still water has traveld 20 km against the current and 180km with the curren

Ira Lisetskai [31]

Answer:

10 kph

Step-by-step explanation:

Find the speed of the current of the river.

20/(20 - c) + 180/(20 + c) = 8

20(20 + c) + 180(20 - c) = 8(400 - c^2)

8c^2 - 160c + 800 = 0

8(c^2 - 20c + 100) = 0

8(c - 10)^2 = 0

c = 10

The speed of the current of the river is 10 km/hour.

5 0

2 years ago

Read 2 more answers

5th grade math. Correct answer will be marked brainliest. Which boxes go where?

loris [4]

Answer:

.46 x 14 = 6.44

46 x 1.4 = 64.4

46 x 14 = 644

4.6 x 14 = 64.4

46 x .14 = 6.44

Step-by-step explanation:

3 0

2 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

Need help ^^^^^^^^^^

NeX [460]

The answer is B. 2y+3
this is proven by multiplying the divisor of the original question and the solution
$(y-5)(2y+3)
 2y^{2}+3y-10y-15
 2y^{2}-7y-15$

4 0

3 years ago

Can I get some help please ​

Can I get some help please