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SSSSS [86.1K]
3 years ago
11

Hannah had some tants 1/5 of the tarts were pineapple tarts and the rest were coconut tarts She gave 1/2 of the coconut tarts to

her uncle and she had 18 coconut tarts left. What was the total number of pineapple and coconut tarts Hannah had at first?​
Mathematics
1 answer:
AnnZ [28]3 years ago
5 0

Answer:

Total tarts = 45

Step-by-step explanation:

Let x be the total tarts

Pineapple tarts = 1/5

So, coconut tarts are 4/5

She gave 1/2 of the coconut tarts to her uncle,

1/2 * 4/5 = 4/10 = 2/5

So, to find x

4/5x - 2/5x = 18

2/5x = 18

2x = 18 x 5

x = 90/2

x = 45

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Find an exact value cos 75°
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Step-by-step explanation:

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Which is the sum of 3.15 × 107 + 9.3 × 106? Write your answer in scientific notation.
BartSMP [9]

Answer:

Scientific Notation:

1.32285 × 10^3

E-Notation:

1.32285e3

Engineering Notation:

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Step-by-step explanation:

7 0
3 years ago
At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel
Alexeev081 [22]
Use Law of Cooling:
T(t) = (T_0 - T_A)e^{-kt} +T_A
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
42 = (80-20)e^{-3k} +20 \\  \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\  \\ k = -\frac{1}{3} ln (\frac{11}{30})

Substituting k back into cooling equation gives:
T(t) = 60(\frac{11}{30})^{t/3} + 20

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\  \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80
Solve for x:
x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80
Sub back into original cooling equation, x = T(t)
-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20
Solve for t:
60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3}  = 60 \\  \\  (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\  \\  \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\  \\ \\  t = 4.959

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
8 0
3 years ago
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