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9966 [12]
3 years ago
7

Which could be the first step in solving the equation represented by the model below? 3 x tiles and 4 negative 1 tiles = 2 x til

es and 5 positive 1 tiles
Mathematics
2 answers:
jonny [76]3 years ago
7 0

Answer: removing 1 x-tile from each side

Step-by-step explanation:

mihalych1998 [28]3 years ago
7 0

Answer:

Its C

Step-by-step explanation:

You might be interested in
Plz help me with this one I don’t get
borishaifa [10]

Answer:

-6x +9

Step-by-step explanation:

You're apparently expected to recognize that equation solution processes generally involve eliminating parentheses using the distributive property. When you do this, you get ...

... (-3)(2x) + (-3)(-3)

... = -6x +9

_____

<em>Comment on the question</em>

Depending on the inequaltiy, you may want to divide by -3 first in your solution process. The question seems to assume there is One Right Way. In math, there is not.

8 0
3 years ago
More series issues!
choli [55]
It can't be that p makes the series converge, because this would introduce a zero in the denominator when n=1. For a similar reason, p=0 would involve an indeterminate term of 0^0.

That leaves checking what happens when p>0. First, consider the function

f(x)=\dfrac{(\ln x)^p}x

and its derivative

f'(x)=\dfrac{p(\ln x)^{p-1}-(\ln x)^p}{x^2}=\dfrac{(\ln x)^{p-1}}{x^2}(p-\ln x)

f(x) has critical points at x=1 and x=e^p. (These never coincide because we're assuming p>0, so it's always the case that e^p>1.) Between these two points, say at c=\dfrac{e^p}2, you have f'(c)=\dfrac{4\ln2}{e^{2p}}(\ln2)^{p-1}, which is positive regardless of the value of p. This means f(x) is increasing on the interval (1,e^p).

Meanwhile, if x>e^p - and let's take c=2e^p as an example - we have f'(c)=\dfrac{(\ln2+p)^{p-1}}{4e^{2p}}(-\ln2)^{p-1}, which is negative for all p>0. This means f(x) is decreasing for all x>e^p, which shows that \dfrac{(\ln n)^p}n is a decreasing sequence for all n>N, where N is any sufficiently large number that depends on p.

Now, it's also the case that for p>0 (and in fact all p\in\mathbb R),

\displaystyle\lim_{n\to\infty}\dfrac{(\ln n)^p}n=0

So you have a series of a sequence that in absolute value is decreasing and converging to 0. The alternating series then says that the series must converge for all p>0.

For the second question, recall that

h_n=\displaystyle\sum_{k=1}^n\frac1k=1+\frac12+\cdots+\frac1{n-1}+\frac1n
s_n=\displaystyle\sum_{k=1}^n\frac{(-1)^{k-1}}k=1-\frac12+\cdots-\frac1{n-1}+\frac1n

(note that the above is true for even n only - it wouldn't be too difficult to change things around if n is odd)

It follows that

h_{2n}=\displaystyle\sum_{k=1}^{2n}\frac{(-1)^{k-1}}k=1+\frac12+\cdots+\frac1{2n-1}+\frac1{2n}
s_{2n}=\displaystyle\sum_{k=1}^{2n}\frac{(-1)^{k-1}}k=1-\frac12+\cdots+\frac1{2n-1}-\frac1{2n}

Subtracting h_{2n} from s_{2n}, you have

\displaystyle s_{2n}-h_{2n}=(1-1)+\left(-\frac12-\frac12\right)+\left(\frac13-\frac13\right)+\left(-\frac14-\frac14\right)+\cdots+\left(\frac1{2n-1}-\frac1{2n-1}\right)+\left(-\frac1{2n}-\frac1{2n}\right)
s_{2n}-h_{2n}=-1-\dfrac12-\cdots-\dfrac2{2n}
s_{2n}-h_{2n}=-\left(1+\dfrac12+\cdots+\dfrac1n\right)
s_{2n}-h_{2n}=-h_n
\implies s_{2n}=h_{2n}-h_n

as required. Notice that assuming n is odd doesn't change the result; the last term in h_{2n} ends up canceling with the corresponding term in s_{2n} regardless of the parity of n.
5 0
4 years ago
Please help me thank you!!!
g100num [7]

Answer:

The answer is D

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Which triangle(s) are similar to ABC?
Harlamova29_29 [7]
Triangle MNP

Sorry if I’m wrong but I think that’s the one
8 0
3 years ago
The equation x + 3y=9 and the equation 2x+6y=18 are plotted on the same graph chart. All of the following points will lie on bot
PolarNik [594]
Im sorry all i know is that the lines are parallel and A and B are definitely on the line, i believe the answer is D
5 0
3 years ago
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