Question

Marks: 4
[10.02] What are the solutions for x in the proportion 2 times x plus 3 all over x equals 2 times x all over 2.?
Choose one answer.
a. x = −1 and x = 3
b. x = 1 and x = −3
c. x = −2 and x = 3
d. x = 2 and x = −3

@wild_rebel_gurl21

Answer 1

\[ \frac{2x+3}{x}=\frac{2x}{2} \, ?\]

Answer 2

Yes, thats the question

Answer 3

The right fraction can me simplified to just \(x\) since the 2's cancel out

\[ \frac{2x+3}{x}=x\] cross multiply to get
\[ 2x+3=x^2\\
x^2-2x-3=0\] can you factor this or use the quadratic formula to find the answer?

Answer 4

(x+6) / 2x = x / 2
2(x+6) = 2x²
-2x²+2x+12 = 0
divide both sides by -2
x²-x-6 = 0
(x-3)(x+2) = 0
x=3
x=-2

Answer 5

so I thinik that the answer is C

Answer 6

how did the numerator go from 2x+3 to x+6?

Answer 7

0,0

Answer 8

\( x^2-2x-3=0\\
(x-3)(x+1)=0\)

Answer 9

So B

Answer 10

not quite

Answer 11

(x - 3) = 0, or (x + 1) = 0
x = 3, or x = -1

Answer 12

no im telling you the answers C im correct

Answer 13

but its ur testand ur call

Answer 14

http://www.wolframalpha.com/input/?i=%282x%2B3%29%2Fx%3D%282x%29%2F2

Answer 15

Wont load for me.

Answer 16

well it also confirms that the solution is x=-1, x=3

Answer 17

So A

Answer 18

yea

Answer 19

Ok bro thank you! :3mindhelping me with a few more?

Answer 20

ok

Answer 21

Thanks!
Marks: 4
[10.03] Which expression is the simplified form of the square root of k to the seventeenth power.?
Choose one answer.
a. k8the square root of k.
b. kthe square root of k to the fourth power.
c. k4the square root of k.
d. k5the square root of k to the seventh power.

Answer 22

Recall that \(\Large \sqrt{x}=x^{\frac{1}{2}}\)
So,\[\Large \sqrt{k^{17}}=\left( k^{17}\right)^{\frac{1}{2}}=k^{\frac{17}{2}}=k^{\frac{16}{2}+\frac{1}{2}}=k^8k^\frac{1}{2}=k^8\sqrt{k}\]

Answer 23

So for the K 17/2 you minus 1 from the top?

Answer 24

well I separated \(\large \frac{17}{2}=\frac{16+1}{2}=\frac{16}{2}+\frac{1}{2}\)

Answer 25

Ahhhh ok

Answer 26

Have time for another one?

Answer 27

sure

Answer 28

Marks: 4
[10.03] What is the simplified form of the square root of the quantity 72 times n to the thirteenth power.?
Choose one answer.
a. 8nthe square root of the quantity 3 times n to the sixth power.
b. 3n6the square root of the quantity 8 times n.
c. 8n6the square root of the quantity 3 times n.
d. 6n6the square root of the quantity 2 times n.

Answer 29

is that \[\Large \sqrt{72n^{13}} \]

Answer 30

Yes

Answer 31

Sorry if i take a couple to answer, had to help my dad

Answer 32

\[\Large \sqrt{72n^{13}}\\~ \\\Large =\sqrt{72}\sqrt{n^{13}}\\ ~ \\ \Large =\sqrt{9\times 4 \times 2}\left(n^{13} \right)^{\frac{1}{2}}\\ ~ \\ \Large =\sqrt{9}\sqrt{4}\sqrt{2}\left(n^{\frac{13}{2}}\right) \\ ~ \\ \Large =3(2)\sqrt{2}\left( n^{\frac{12}{2}}n^{\frac{1}{2}}\right)\\
\Large =6n^6\sqrt{2n}\]

Answer 33

Thanks man! Can u be like my tutor lol

Answer 34

your welcome :) hehe well you can tag me in a question if you need it ;)

Answer 35

Well i have a few more if your not to busy

Answer 36

Can you post it in a new question? It's getting quite long the post here, and the more formulas posted, the more it lags :\