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Softa [21]
3 years ago
5

Please help !! Geometry homework. Will mark brainliest answer !

Mathematics
1 answer:
mr Goodwill [35]3 years ago
8 0

Answer:

m\widehat{BD}=70^{\circ}

Step-by-step explanation:

APB is a straight line.

Using liner pair(sum of angles on straight line is 180 degree)

\angle APF+\angle FPB=180^{\circ}

But \angle APF=110^{\circ}

\therefore 110^{\circ|+\angle FPB=180^{\circ}

\angle FPB=70^{\circ}

In triangle FPD, FP=DP (radius of same circle.

So, triangle FPD would be isosceles triangle. FD is base of triangle.

PE is perpendicular to FD.

As we know angle bisector of isosceles triangle make perpendicular to base.

So, line PE bisect angle FPD.

\therefore \angle FPE=\angle DPE

As we found above \angle FPE=70^{\circ}

So, \angle DPE=70^{\circ}

m\widehat{BD}=70^{\circ}

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Given

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3 years ago
The daily average temperature in Santiago, Chile, varies over time in a periodic way that can be modeled approximately by a trig
natali 33 [55]

Answer:

a) the trigonometric function is;

y = 7.5 sin ( \frac{2 \pi}{365}t + \frac{337 \pi}{730})+ 21.5

b) y = 28.36^0 \ C    ( to two decimal places)

Step-by-step explanation:

This data can be represented by the sinusoidal function of the form :

\mathbf{y = A sin (Bt -C)+D}

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A = \dfrac{29-14}{2}

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D = \frac{29+14}{2}

D = 21.5

Replacing the values of A ; B and D into the above sinusoidal function; we have :

y = 7.5 sin (\frac{2 \pi}{365}t -C) + 21.5

From the question; when it is 7th of the year ( i.e January 7);

t =  7 and the temperature (y) = 29° C

replacing that too into the above equation; we have:

29= 7.5 sin (\frac{2 \pi}{365}*7 -C) + 21.5

29= 7.5 sin (\frac{14 \pi}{365} -C) + 21.5

\frac{29-21.5}{7.5}=  sin (\frac{14 \pi}{365} -C)

1=  sin (\frac{14 \pi}{365} -C)

sin^{-1}(1)=   (\frac{14 \pi}{365} -C)

\frac{\pi}{2}=   (\frac{14 \pi}{365} -C)

C=   (\frac{14 \pi}{365} -\frac{\pi}{2})

C=   (\frac{28 \pi- 365 \pi}{730} )

C=  \frac{-337 \pi}{730}

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y = 7.5 sin ( \frac{2 \pi}{365}t + \frac{337 \pi}{730})+ 21.5

Similarly; to determine the temperature o Jan 31; i.e when t= 31 ; we have :

y = 7.5 sin ( \frac{2 \pi}{365}*31+ \frac{337 \pi}{730})+ 21.5

y = 7.5 sin ( \frac{62 \pi}{365}+ \frac{337 \pi}{730})+ 21.5

y = 7.5 sin ( \frac{124 \pi+ 337 \pi }{730})+ 21.5

y = 7.5 sin ( \frac{461 \pi }{730})+ 21.5

y = 7.5 *( 0.915)+ 21.5

y = 6.8689+ 21.5

y = 28.36^0 \ C    ( to two decimal places)

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