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4 years ago

Question 2 When writing music, musicians write the music on paper instead of notebook paper.

Mathematics

2 answers:

n200080 [17]4 years ago

4 0

Answer:

true ( i think its been long time)

Step-by-step explanation:

I am Lyosha [343]4 years ago

4 0

Answer:

musicians write there music on whatever they can find so this would be false

Hope this helps^^

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Jennifer is taking a quiz and has answered 13 questions so far. if there are 25 questions on Jennifer's test, how many questions

tresset_1 [31]

Answer:

12 questions left.

Step-by-step explanation:

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2 years ago

Calculate the slope between the points: (4, -9) and (5, 1)

12345 [234]

Answer:

The answer is A- m=10

Step-by-step explanation:

The way to find slope is known as "Rise over Run" meaning the increase of Y over the increase of X. The increase of Y is 10 because 1 - (-9) = 10. The increase of X is 1 because 5 - 4 = 1. 10 over 1, or 10 / 1, equals 10.

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4 years ago

The equation y= x-3 describes how the variables x and y are related

OLga [1]

Answer:

The amount y is depends on what x's value is. It is helpful to make a table or graph so we can see how the y value changes like the x value :)

Have an amazing day!!

Please rate and mark brainliest!!

8 0

2 years ago

Read 2 more answers

Will give brainliest

solong [7]

Answer:

idk

Step-by-step explanation:

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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

4 years ago