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3 years ago

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What are the names of three planes that contain point B? a planes ABD, EFG, and BDH b planes ABD, AEF, and DHF c planes CDH, ABF

, and BDH d planes ABD, ABF, and CDH

1 answer:

Radda [10]3 years ago

8 0

Answer

Sol: The correct answer is b. three planes that contain point B are ABD, AEF and DHF.

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Find the six trig function values of the angle 240*Show all work, do not use calculator

-BARSIC- [3]

Solution:

Given:

$240^0$

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

$sin240^0=sin(180+60)$

Using the trigonometric identity;

$sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny$

Hence,

$\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}$

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

$cos240^0=cos(180+60)$

Using the trigonometric identity;

$cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny$

Hence,

$\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\ \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}$

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

$tan240^0=tan(180+60)$

Using the trigonometric identity;

$tan(180+x)=tan\text{ }x$

Hence,

$\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\ \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}$

To get cosec 240 degrees:

$\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}$

To get sec 240 degrees:

$\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\ \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\ \\ Thus, \\ sec240^0=-2 \end{gathered}$

To get cot 240 degrees:

$\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\ \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}$

5 0

7 months ago

What’s the correct answer for this?

slava [35]

Answer:

1/4

Step-by-step explanation:

Let's denote the probabilities as following:

Probability that a teenager has a sister:

P(A) = 12/28

Probability that a teenager has a brother:

P(B) = 7/28

Probability that a teenager has both a sister and a brother:

P(A⋂B) = 3/28

Probability that a selected teenager has a sister also has a brother, or in other words, he/she has a brother, given he/she had a sister:

P(B|A)

Let's apply the formula of conditional probability to work out P(B|A)

P(B|A) = P(A⋂B)/P(A) = (3/28)/(12/28) = (3*28)/(12*28) = 3/12 = 1/4

=> Option C is correct

Hope this helps!

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Please answer correctly !!!!!!!!!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!!!

Andreyy89

Answer:

I don't have a protractor with me but if you do look where x is and line the little hole on the bottom of the protractor on the crossing point and see what degree the angle is at. Hope I helped :)

Step-by-step explanation:

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