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natima [27]
3 years ago
10

Write your answer anywhere in the open area of each question

Mathematics
1 answer:
Leto [7]3 years ago
5 0
I dont knowwwee....kaksk
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What is 16.2 simplified (and 121 simplified to)
Cerrena [4.2K]

Answer:

16.2 = \frac{81}{5} and 121 = \frac{121}{1}

Step-by-step explanation:

16.2 can be simplified to its fraction form, which is \frac{81}{5}

  • 16+\frac{2}{10} =\frac{81}{5} which is 16.2

121 is already simplified since it is a whole number, but its fraction form would be \frac{121}{1}

  • 121+\frac{0}{10} =121 which is 121

-----

<em>(I hope this helped & good luck! <3)</em>

7 0
3 years ago
16.8 is 10% of what number? Someone help TWT
Aloiza [94]

Answer:

168

Step-by-step explanation:

16.8 = 10%

x = 100%

To get to 100%, you have to multiply 10 by ten and 16.8 by 10. Like this:

16.8 = 10%                   original equation

16.8 (10) = 10% (10)      * both sides by 10

168 = 100%                  simplify

:3))

7 0
3 years ago
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Please help!!!
IrinaK [193]

Answer:

it would be $16 off 40 so the pants would be $24

3 0
3 years ago
WILL GIVE BRAINLIEST TO RIGHT ANSWER!!
daser333 [38]

Answer:

Step-by-step explanation:

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3 0
3 years ago
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Given sin(−θ)=1/5 and tanθ=√6/12 .
Natasha_Volkova [10]

Answer:

B). −2√6/5

Step-by-step explanation:

tan theta = sin theta/ cos theta

Multiply each side by cos theta

tan theta * cos theta = sin theta

Divide each side by tan theta

cos theta = sin theta/ tan theta

We know that the sin (- theta) = - sin theta  since sin is and odd function

sin theta  = - (  sin (-theta))

Putting this into the above equation,

cos theta =  - (  sin (-theta)) / tan theta

cos theta = - 1/5 / (sqrt(6)/12)

Remember when dividing fractions, we use copy dot flip

cos theta =   -1/5  * 12/ sqrt(6)

cos theta = -12/ (5 sqrt(6))

We cannot leave a sqrt in the denominator, so multiply the top and bottom by sqrt(6)/sqrt(6)

cos theta = -12/ (5 sqrt(6))  * sqrt(6)/sqrt(6)

cos theta = -12 sqrt(6) / 5*6

Simplify the fraction.

cos theta = -2 sqrt(60/5



3 0
3 years ago
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