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2 years ago

The correct answer is a right?

Mathematics

1 answer:

svetoff [14.1K]2 years ago

5 0

We can see that a and b are parallel, and c and e are parallel, so the correct option is E.

<h3>Which line must be parallel?</h3>

On the diagram, we can see that the angles in the third quadrant of the intersections between a and c, and the intersections between b and c, are the same angle.

Then, lines a and b must be parallel.

For the intersections with line d, we can see that this time the angle is on the fourth quadrant, so c and d are not parallel.

Finally, for line e, we can see that the known angle is on the first quadrant.

Notice that the angle on the first quadrant will be equal to the angle on the third quadrant.

So for the intersections of a and e, and b and e, on the third quadrant we have the known angle (the same one as in the intersections of a and c, and a and b).

Then c and e are parallel.

Then A and C are true.

Thus, the correct option is E.

If you want to learn more about parallel lines:

brainly.com/question/24607467

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Ahhh okay help me pls!

inna [77]

Answer:

The answer is Dilation of a scale factor of 4 followed by a reflection of the y axis.

Step-by-step explanation:

On the smaller triangle, the line EF has a length of 1 unit.

On the larger triangle, the line EF has a length of 4 units.

Therefore, the smaller triangle was dilated by 4, since 4x1 is 4.

The triangles was flipped over the y axis.

This means that it was reflected over the y axis.

4 0

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:

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