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baherus [9]
3 years ago
15

Consider the points.

Mathematics
1 answer:
miskamm [114]3 years ago
3 0

Answer:

6 units

Step-by-step explanation:

Use the distance formula, \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}.

Plug the x and y values from the given points into the distance formula:

\sqrt{([-4]-2)^{2}+(5-5)^{2}}.

Solve the square root:

\sqrt{36} = \sqrt{6^{2}} = 6

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‼️ IM BEING TIMED PLEASE HELP‼️<br><br> Which is the graph of fix) = 4(3)?
dexar [7]

Answer:

Step-by-step explanation:

After graphing the function f(x) = 4(\frac{1}{2})^{x}  we can see that the correct graph is the middle one in the picture. This is because with an x-input of 1 the value of 1/2 will stay the same and when multiplied by 4 it would give us a y-coordinate of 2 as seen in the middle graph. The y-coordinate keeps decreasing exponentially by half for every increase in the x-coordinate. A zoomed-in version of the graph can be seen in the attached picture below for a better understanding.

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3 years ago
In the figure, angle E measures 30°, angle F measures 57°, and angle B measures 27°. What is the measurement of angle A? 27° 66°
Thepotemich [5.8K]
It would be 66 degrees because

this would equal 180 degrees

add the angles together:30+57+27=114

now subtract 180-114=66degrees

FINAL ANSWER:66degrees 
3 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
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4 years ago
Many golfers wear wrist bracelets containing magnets because they claim the magnets improve balance and the length of shots play
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I think my answer is a because it can’t be B and it can’t be C and it can’t be D so my answer is a
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3 years ago
Read 2 more answers
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