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erastovalidia [21]
4 years ago
12

Which expression has a pair of like terms?

Mathematics
1 answer:
KATRIN_1 [288]4 years ago
7 0

Answer: x-3+x+3y

Step-by-step explanation:

Because there is 2 x’s

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PLZ HELP HELP!!!!!!!Two pairs of jeans cost $75 before tax. If the total paid for the two pairs of jeans with tax was $81, what
Anna71 [15]
8%
6 divided by 75 is .08.
3 0
3 years ago
Please help I have a sucky teacher and don’t know what to do
Katyanochek1 [597]

Answer:

  1, 4, 3, 6, 2, 7, 5

Step-by-step explanation:

The circle operator in f\circ g means that the function named on the left (f) operates on the result of the function named on the right (g). So, the notation

  (f\circ g)(x)

means

  f(g(x))

That is, g(x) is computed and used as input to the function f. When the function definitions are as shown, the evaluation proceeds in these steps.

  (f\circ g)(-4)\\\\=f(g(-4))\quad\text{write as function of a function}\\\\=f(-2(-4)^2)\quad\text{put -4 for x in $g(x)$}\\\\=f(-2(16))\quad\text{evaluate $(-4)^2$}\\\\=f(-32)\quad\text{evaluate $g(-4)$}\\\\=2(-32-1)\quad\text{put -32 for x in $f(x)$}\\\\=2(-33)\quad\text{first step evaluating $f(-32)$}\\\\=-66\quad\text{value of $(f\circ g)(-4)$}

__

If your step choices at the bottom are numbered 1 to 7 left-to right, then the sequence is as shown in the Answer section above.

4 0
3 years ago
Choose a method . Then find the product 10x15
34kurt

Answer:

150

Step-by-step explanation:

15*10

Lets get rid of the 0 for now

15*1

That is 15 added to itself 0 times so it is 15

Now lets put the 0 back

and we will get

150

5 0
3 years ago
Read 2 more answers
Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini
Nat2105 [25]
By inspection, it's clear that the sequence must converge to \dfrac32 because

\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32

when n is arbitrarily large.

Now, for the limit as n\to\infty to be equal to \dfrac32 is to say that for any \varepsilon>0, there exists some N such that whenever n>N, it follows that

\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|

From this inequality, we get

\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}
\implies|2n-1|>\dfrac5{2\varepsilon}
\implies2n-1\dfrac5{2\varepsilon}
\implies n\dfrac12+\dfrac5{4\varepsilon}

As we're considering n\to\infty, we can omit the first inequality.

We can then see that choosing N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that N\in\mathbb N.
6 0
3 years ago
Determine the number of Solutions for the system of equations y = 2x + 3 and y = 2x +3
vampirchik [111]

Answer:

:/

Step-by-step explanation:

:/

7 0
3 years ago
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