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Elanso [62]
3 years ago
8

What is the value of x? х= 32 х= 36 х = 37 x= 40

Mathematics
2 answers:
labwork [276]3 years ago
8 0

Answer:

x = 40 deg

Step-by-step explanation:

Given that the line at the base of the triangle is a continuous straight line,

x + (4x-20) = 180 degrees

x + 4x  - 20 = 180

5x = 180 + 20

5x = 200

x = 40 deg

Akimi4 [234]3 years ago
7 0
The answer is 40 because 5x=200 and so X must equal 40
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A DVD was priced at $2.87 last week. This week the same DVD is priced at $10.59. What is the percent increase in price?
solniwko [45]

Answer:

Percentage increase = 269%

Step-by-step explanation:

Price of a DVD last week = $2.87

Price of the same DVD this week = $10.59

Increase in the price of DVD between last week and this week = 10.59 - 2.87

= $7.72

Therefore, percentage increase in the price of DVD = \frac{\text{Increase in the price}}{\text{Price of DVD last week}}\times 100

= \frac{7.72}{2.87}\times 100

= 268.99

= 269%

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3 years ago
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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
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4 years ago
Kyle built a tree house 4 ft. by 6 ft. What was the area of the tree house?
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Step-by-step explanation:

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