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raketka [301]
3 years ago
11

Mrs. Johnson put a fence around the basil. If she fence is 2 ft from the edge of the garden on each side, what is the perimeter

of the fence in feet?
Mathematics
2 answers:
Vadim26 [7]3 years ago
7 0

The perimeter of the fence is 23 1/2 ft

natka813 [3]3 years ago
6 0
The perimeter would be 8ft because 2+2+2+2= 8
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A submarine moved from the water surface at a speed of -20 m per minute . How far is the
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Step-by-step explanation:

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To the nearest hundredth, what is the sine of the acute angle formed by the line whose equation is y = 3x - 3 and the positive x
Serhud [2]

Answer:

The sine of the acute angle is 0.95

Step-by-step explanation:

using a graphing tool

see the attached figure

Let

A-----> the acute angle formed by the given line and the positive x-axis

In the right triangle XYZ

we know that

sin(A)=\frac{xy}{xz}

we have

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yz=1\ units

Applying the Pythagoras Theorem find the value of xz

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5 0
4 years ago
What is 629,462 rounded to the nearest hundred
WITCHER [35]
629,500 is the answer to your question
5 0
4 years ago
Read 2 more answers
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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3 years ago
Eric paints 8 rooms in 3 days for 600$ about how many rooms could Eric paint in 6 days?
Marianna [84]

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16 rooms

Step-by-step explanation:

8 0
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