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Rom4ik [11]
3 years ago
8

15. Aroon has 180 cat toys to pack in boxes. He packs 30 toys in each box. How many boxes does he need? Write an equation using

the letter n to stand for the unknown factor. Explain how to find the unknown factor??
Mathematics
1 answer:
Volgvan3 years ago
7 0

Answer:

n=6

Step-by-step explanation:

n= the amount of boxes he needs

so if he puts 30 toys in each box out of 180 you divide 180 by 30 and you get your answer.

so the amount of boxes he needs = 6

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What is the trigonometric ratio for sin C ?<br><br> Enter your answer, as a simplified fraction.
Contact [7]
By definition we have to:
 Sin (x) = C.O / h
 Where,
 x = angle
 C.O = opposite leg
 h: hypotenuse.
 Substituting values we have:
 Sin (C) = AB / 82
 We should look for AB. For this, we use the following relationship:
 AB ^ 2 + 80 ^ 2 = 82 ^ 2
 Clearing AB:
 AB = root ((82 ^ 2) - (80 ^ 2))
 AB = 18
 Substituting we have:
 Sin (C) = 18/82
 Simplifying:
 Sin (C) = 9/41

 Answer:
 
the trigonometric ratio for sin C is:
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5 0
3 years ago
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aniked [119]

Answer:

$1.98

Step-by-step explanation:

$24.00 multilpy by 8.25%= $1.98. Then you add $24.00 + $1.98 + $25.98

The sales tax is $1.98

4 0
3 years ago
Read 2 more answers
58 1/4 as a decimal
sleet_krkn [62]

Answer:

58.25

Step-by-step explanation:

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8 0
2 years ago
adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the
Zigmanuir [339]

Answer:

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659 and \left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

Step-by-step explanation:

The equation of the isotope decay is:

\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

\tau = \frac{5568\,years}{\ln 2}

\tau \approx 8032.926\,years

The decay time is:

t = 1315\,years + 2007\,years \pm 13\,years (There is no a year 0 in chronology).

t = 3335 \pm 13\,years

Lastly, the relative amount is estimated by direct substitution:

\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659

\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

4 0
3 years ago
EXPLAIN YOUR ANSWER
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Solve the equation by substitution method.
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