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3 years ago

1/5 + 5/13 - 3/14 = ?

Mathematics

1 answer:

forsale [732]3 years ago

3 0

Answer:

$\frac{337}{910}$

Step-by-step explanation:

$\frac{1}{5}+\frac{5}{13}-\frac{3}{14}$

Lowest Common Multiple of 5, 13 and 14 = 910

Convert the fractions based on the denominator:

$\frac{182}{910}+\frac{350}{910}-\frac{195}{910}$

Combine the fractions since the denominator is the same:

$\frac{182+350-195}{910}$

$\frac{337}{910}$

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Mrs.Anderson bought party favors for the 24 students in her class. Each favor cost $2.27. How much did all the party favors cost

laiz [17]

$54.48 because 24 x 2.27 = 54.48

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4 years ago

Which graph represents the solution to the given system? Y=-x+5 y= 1/4 x+10

jeka57 [31]

Answer:

See attachment

Step-by-step explanation:

The given system is :

$y = - x + 5$

$y = \frac{1}{4}x + 10$

Let us equate both equations and solve for x.

$- x + 5 = \frac{1}{4}x + 10$

$- 4x + 20 = x + 40$

$20 - 40 =x + 4x$

$- 20 = 5x$

$x = - 4$

$y = - - 4 + 5 = 9$

The the two lines of this system will meet at (-4,9)

The graph is shown in the attachment.

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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

3 years ago

PLEASE HELP -7y + 11 = 75 + y

Masja [62]

Answer:

y = -8

Step-by-step explanation:

-7y + 11 = 75 + y

Bring the "y" variable on one side, and rest on the other.

Rearranging, we get,

-7y - y = 75 - 11

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-y = 8

y = -8

Hope it helps :)

3 0

2 years ago

Since it is a. repeating decimal 14.1 is irrational. True or false statement

tatyana61 [14]

False this statement is false

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3 years ago

Read 2 more answers