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4 years ago

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