Question

simplify the expression.

Answer 1

\[\sqrt{3}-4\sqrt{3}\]

Answer 2

same as 1(a)-4(a), because the roots are the EXACTLY the same, they're treated as 1 literal per se

Answer 3

\[-3\sqrt{3}\]

Answer 4

yes

Answer 5

\[4\sqrt{5}+5\sqrt{45} = 19\sqrt{5}\]

Answer 6

yep

Answer 7

\[5=\sqrt{k}+4=1\]

Answer 8

yes, because \(1=\sqrt{k}\text{ and }1^2=(\sqrt{k})^2\)

Answer 9

The formula \[v=\sqrt{64h}\] can be used to find the velocity v in feet per second of an object that has fallen h feet. Find how far the object has fallen if its velocity is 30 feet per second. Round your answer to the nearest hundredth.

Answer 10

V=30feet and thus \( \implies 30 = \sqrt{64h}\)

Answer 11

14.06

Answer 12

feet

Answer 13

yes

Answer 14

solve the equation and identify any extraneous solutions.

\[q=\sqrt{-7q}\]

Answer 15

0 and –7 are solutions of the original equation.
0 is a solution of the original equation. –7 is an extraneous solution.
–7 is a solution of the original equation. 0 is an extraneous solution.
7 is a solution of the original equation. 0 is an extraneous solution.

Answer 16

if you plug in -7 to q, that is q=-7, you'd have \(\implies -7 = \sqrt{-7*-7} \implies -7 = -7\)
which is correct
if you plug in 0 to q, that is \(\implies -7 = \sqrt{-7*0} \implies -7 = 0\), which would make it extraneous

Answer 17

so
-7 is a solution of the original equation. 0 is an extraneous solution.

Answer 18

wait.... shoot, not it wouldn't :(

Answer 19

$$ \implies 0 = \sqrt{-7*0} \implies 0 = 0$$

Answer 20

yes ^^ that's correct..

So is the answer still -7 is a solution of the original equation. 0 is an extraneous solution?

Answer 21

0 and –7 are solutions of the original equation.

Answer 22

oh yeah that's right lol

Answer 23

they both yield a valid EQUATION, both factors on each side of the = sign, correlate

Answer 24

\[y=4\sqrt{2x-5}= x \ge \frac{ 2 }{ 5}\]

Answer 25

did you mean \(x\le\cfrac{5}{2}\)?

Answer 26

\(x\ge\cfrac{5}{2}\) rather

Answer 27

yes i keep messing up haha

Answer 28

if x becomes less than that, the root becomes negative, and thus an 'imaginary' number

Answer 29

wait, no i meant \[x \ge \frac{ 5 }{ 2 }\]

Answer 30

if the DOMAIN for "x" is 5/2 and up, the RANGE is 0..\(\alpha\)

Answer 31

my choices:

\[x \ge \frac{ 5 }{ 2}\]
\[x \ge \frac{ 2 }{ 5 }\]
\[x \ge-\frac{ 5 }{ 2 }\]
\[x >-\frac{ 5 }{ 2 }\]

Answer 32

well, I guess you have the answer already then :)

Answer 33

if x becomes \(\le\cfrac{5}{2}\) that, the root becomes negative, and thus an 'imaginary' number

Answer 34

the first one right? haha
just making sure ;)