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Aleksandr [31]
3 years ago
14

Simplify the expression. 5 + 8x + 3x -2

Mathematics
2 answers:
Alla [95]3 years ago
6 0

Answer: 3 + 11x

Step-by-step explanation: combine like terms 5 -2 is 3 and 8x + 3x is 11x and they're both positive so it is 3 + 11x

stepladder [879]3 years ago
4 0

Answer:

3+11x

Step-by-step explanation:

5+8x+3x-2

8x+3x+5-2

11x+3

3+11x

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Milo has 2 hours of free time. He spends 2/4 of an hour with his dog he spends 3/4 of an drawing. What fraction of an hour dose
TEA [102]

Answer:

3/4 time left

Step-by-step explanation:

7 0
3 years ago
A florist uses 10 red roses for every 4 white roses in her bouquets. What is the ratio of red roses to the total number of flowe
Kruka [31]
10:14 or 10 to 14

The total number of roses is 14 and there are 10 red roses. For every 14 roses there are 10 red roses.
5 0
3 years ago
Read 2 more answers
adult tickets to a play cost $22. tickets for children cost $15. tickets for a group of 11 people cost a total of $228. write an
Tom [10]

Adults = 9

Children = 2


Let x be the number of adults.

11-x is the number of children.


22x + 15(11-x) = 228

22x + 165 - 15x = 228

22x - 15x = 228 - 165

7x = 63

7x / 7 = 63 / 7

x = 9  number of adults.


11 - x = 11 - 9 = 2 number of children.


To check:


22x + 15(11-x) = 228

22(9) + 15(11-9) = 228

198 + 30 = 228

228 = 228


(not my answer btw)

3 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
What is the value of the expression below?<br> -8 + 19 + 8<br> 0<br> 11<br> 19<br> 35
Inga [223]

Answer:

19

Step-by-step explanation:

-8+8=0

0+19=19

Hope it helps.

8 0
3 years ago
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