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4 years ago

< Salir

Mathematics

1 answer:

fenix001 [56]4 years ago

6 0

Answer:

Option C.

Step-by-step explanation:

Kurt started for the school at 07:35 and ends at 08:25.

Time taken by Kurt to reach the school

= 08:25 - 07:35

= (07 hours + 85 minutes) - (07 hours + 35 minutes) [Since 08 hours + 25 minutes = 07 hours + (60 + 25)minutes]

= 85 minutes - 35 minutes

= 50 minutes

Time taken by Drew to reach the school

= 08:30 - 07:55

= (07 hours + 90 minutes) - (07 hours + 55 minutes) [Since 08 hours + 30 minutes = 07 hours + (60 + 30)minutes]

= 90 minutes - 55 minutes

= 35 minutes

Difference in the time taken by Kurt and Drew = 50 - 35 = 15 minutes.

Therefore, Drew is 15 minutes faster than Kurt.

Option C. is the answer.

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Answer:

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Step-by-step explanation:

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3 years ago

Need help with trig problem in pic

Sidana [21]

Answer:

a) $cos(\alpha)=-\frac{3}{5}\\$

b) $\sin(\beta)= \frac{\sqrt{3} }{2}$

c) $\frac{4+3\sqrt{3} }{10}\\$

d) $\alpha\approx 53.1^o$

Step-by-step explanation:

a) The problem tells us that angle $\alpha$ is in the second quadrant. We know that in that quadrant the cosine is negative.

We can use the Pythagorean identity:

$tan^2(\alpha)+1=sec^2(\alpha)\\(-\frac{4}{3})^2 +1=sec^2(\alpha)\\sec^2(\alpha)=\frac{16}{9} +1\\sec^2(\alpha)=\frac{25}{9} \\sec(\alpha) =+/- \frac{5}{3}\\cos(\alpha)=+/- \frac{3}{5}$

Where we have used that the secant of an angle is the reciprocal of the cos of the angle.

Since we know that the cosine must be negative because the angle is in the second quadrant, then we take the negative answer:

$cos(\alpha)=-\frac{3}{5}$

b) This angle is in the first quadrant (where the sine function is positive. They give us the value of the cosine of the angle, so we can use the Pythagorean identity to find the value of the sine of that angle:

$cos (\beta)=\frac{1}{2} \\\\sin^2(\beta)=1-cos^2(\beta)\\sin^2(\beta)=1-\frac{1}{4} \\\\sin^2(\beta)=\frac{3}{4} \\sin(\beta)=+/- \frac{\sqrt{3} }{2} \\sin(\beta)= \frac{\sqrt{3} }{2}$

where we took the positive value, since we know that the angle is in the first quadrant.

c) We can now find $sin(\alpha -\beta)$ by using the identity:

$sin(\alpha -\beta)=sin(\alpha)\,cos(\beta)-cos(\alpha)\,sin(\beta)\\$

Notice that we need to find $sin(\alpha)$ , which we do via the Pythagorean identity and knowing the value of the cosine found in part a) above:

$sin(\alpha)=\sqrt{1-cos^2(\alpha)} \\sin(\alpha)=\sqrt{1-\frac{9}{25} )} \\sin(\alpha)=\sqrt{\frac{16}{25} )} \\sin(\alpha)=\frac{4}{5}$

Then:

$sin(\alpha -\beta)=\frac{4}{5}\,\frac{1}{2} -(-\frac{3}{5}) \,\frac{\sqrt{3} }{2} \\sin(\alpha -\beta)=\frac{2}{5}+\frac{3\sqrt{3} }{10}=\frac{4+3\sqrt{3} }{10}$