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SashulF [63]
3 years ago
12

A skier has decided that on each trip down a slope, she will do 3 more jumps than before. On her first trip she did 5 jumps. Der

ive the sigma notation that shows how many total jumps she attempts from her third trip down the hill through her tenth trip. Then solve for the number of total jumps from her third to tenth trips.
Mathematics
1 answer:
Evgesh-ka [11]3 years ago
6 0
Your answer is 32 hope it helped
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Jay has a budget of $100. Jay spent $13 on movie tickets, $38 on dinner with his friend, and $11 on a taxi ride home. How much m
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38

Step-by-step explanation:

100-13-38-11=38

So Jay has 38 dollars on the budget

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Mrs. Bolch has been in martial arts for 85% of her life. She’s almost 35. How long has she been in martial arts (round to neares
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Ryan and Kaine are two partners in a business. Ryan makes $3 in profits for every $4 Kaine makes. If Kaine makes $60 profit on t
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Answer is $45

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6 0
3 years ago
GEOMETRY Help !! Which term describes the red curve in the figure below?
AleksandrR [38]
It is an elipse 
D is the answer
7 0
3 years ago
A sample of tritium-3 decayed to 94.5% of its original amount after a year.
Zolol [24]
(a) If y(t) is the mass after t days and y(0) = A then y(t) = Ae^{kt}.

y(1) = Ae^k = 0.945A \implies e^k = 0.945 \implies k = \ln 0.945

\text{Then } Ae^{(\ln 0.945)t} = \frac{1}{2}A\ \Leftrightarrow\ \ln e^{(\ln 0.945)t} = \ln \frac{1}{2}\ \Leftrightarrow\  ( \ln 0.945) t = \ln \frac{1}{2} \Leftrightarrow \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 12.25 \text{ years}

(b)
Ae^{(\ln 0.945) t} = 0.20 A \ \Leftrightarrow\ (\ln 0.945) t \ln \frac{1}{5}\ \implies \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 28.45\text{ years}
3 0
3 years ago
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