Question

http://prntscr.com/lnnio2

Answer 1

@Vocaloid

Answer 2

@Hero

Answer 3

For the top one you just need to re-write both terms by moving the decimal point the appropriate number of spaces indicates by the exponent

For example, 1.26 * 10^(-3) means take 1.26 and move the decimal place to the left three places to get standard form

Answer 4

For the bottom one you just need for apply the exponent rule to the product, remember that when you multiply the same base together you’d add the exponents

Answer 5

the first one is D?

Answer 6

Yes

Answer 7

and the second im also thinking D or B

Answer 8

Remember you just add the exponents and keep the base 10

Answer 9

but would we add or multiply 4 and 5?

Answer 10

Multiply

Answer 11

so A

Answer 12

Yes

Answer 13

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Answer 14

for the top one use the distance formula; for the second one use the midpoint formula

Answer 15

wouldnt it be \[\sqrt{40}\]?

Answer 16

or am i doing it wrong?

Answer 17

check your calculations again, be careful with signs

Answer 18

isnt it like 3.16 something?

Answer 19

since the results are in radical form, leave your solutions in radical form

Answer 20

sqrt[(-2-1)^2 + (4-3)^2] = ?

Answer 21

but thats the decimal so the radical form is \[\sqrt{10}\]

Answer 22

yup that's it

Answer 23

and the second one is B?

Answer 24

yes

Answer 25

http://prntscr.com/lnoges

Answer 26

pythagorean theorem

a^2 + b^2 = c^2

where c is the hypotenuse, and a and b are the other two sides

you are given c = 5sqrt(2) (the hypotenuse)
and a = 5 (one of the other sides)
so you can solve for b (the last side)

Answer 27

like this?
http://prntscr.com/lnoinc

Answer 28

that final exponent 2 needs to apply to the entire expression

[ 5sqrt(2) ] ^2

Answer 29

i keep getting b=15

Answer 30

5^2 + b^2 = [5sqrt(2)]^2
if all else fails try solving it manually

Answer 31

b=5

Answer 32

good

Answer 33

are these correct
http://prntscr.com/lnolj3

Answer 34

yup

Answer 35

http://prntscr.com/lnom7t

Answer 36

good

Answer 37

okay tyy i got them all right