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barxatty [35]
2 years ago
11

50 POINTS Vanessa has 6 dimes and 2 pennies. Joachim has 1 dollar, 3 dimes, and 5 pennies. Jimmy has 5 dollars and 7 pennies. Th

ey want to put their money together to buy a game that costs $8.00. Do they have enough money to buy the game? If not, how much more money do they need? SHOW WORK
Mathematics
2 answers:
yKpoI14uk [10]2 years ago
7 0

Answer:they need more money

Step-by-step explanation: they need 96 more cents

Yakvenalex [24]2 years ago
3 0

Vanessa: a dime = 10 cents = 0.10

A penny = 1 cent = 0.01

6 x 0.10 = 0.60

2 x 0.01 = 0.02

Total = 0.60 + 0.02 = 0.62

Joachim : a dollar = 1.00

3 x 0.10 = 0.30

5 x 0.01 = 0.05

Total = 1.00 + 0.30 + 0.05 = 1.35

Jimmy: total = 5.00 + 0.07 = 5.07

Total of all 3: 0.62 + 1.35 + 5.07 = $7.04

This is less than $8 so they don’t have enough.

They need: 8.00 - 7.04 = $0.96 more

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Answer:

-1\frac{7}{10}

Step-by-step explanation:

  1. Convert the mixed numbers into improper fractions: \frac{-17}{8} ÷ \frac{5}{4}
  2. Apply the fractions formula for division: \frac{-17*4}{8*5} = \frac{-68}{40}
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maxonik [38]

Answer with Step-by-step explanation:

We are given that

r(t)=< t^2,\frac{2}{3}t^3,t >

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r'(t)=

\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1

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Now, substitute t=1

T(1)=\frac{}{2+1}=\frac{1}{3}

T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}

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T'(1)=\frac{1}{9}=

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N(1)=\frac{T'(1)}{\mid T'(1)\mid}

N(1)=\frac{}{\frac{2}{3}}=

N(1)=

B(1)=T(1)\times N(1)

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B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k

B(1)=\frac{1}{3}

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