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pshichka [43]
3 years ago
8

If ∠a = 90°, find the measures of ∠b, ∠c and ∠d.

Mathematics
2 answers:
nydimaria [60]3 years ago
5 0

Answer:

m∠b = 35°

m∠c = 145°

m∠ d = 35°

Step-by-step explanation:

To find m∠b, we use Definition of Supplementary Angles:

180 - (90 + 55) = 35°

To find m∠c, we also use Definition of Supplementary Angles:

180 - 35 = 145°

To find m∠d, we use Vertical Angles Theorem:

m∠b = m∠d = 35°

Darya [45]3 years ago
3 0

Answer:

Step-by-step explanation:I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me...

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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
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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
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Substituting k back into cooling equation gives:
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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:
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This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
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First you multiply the numbers you do have to get the number your looking for 1/7•14=2 and 2divided by 1/7=14 so 2 would be your answer
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