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Dvinal [7]
3 years ago
10

98 points people! So answer it plz.

Mathematics
2 answers:
Yanka [14]3 years ago
5 0

Answer:

A. Angle 4 and Angle 12

Step-by-step explanation:

the given lines are parallel so,

angle 8 is corresponding to angle 4 and angle 12,

so, the required answer is A.

Arisa [49]3 years ago
4 0
I think the answer to this question is A
You might be interested in
Hope u can read it sorry
Crank
B. 2x5square root of 3
4 0
4 years ago
Read 2 more answers
Suppose in Loma, Montana the temperature rose from -2.3 degrees Fahrenheit to 8.6 degrees Fahrenheit. What is the change in temp
adelina 88 [10]

Answer:

A: 10.9

Step-by-step explanation:   -2.3 is 2.3 away from zero then add the length from zero to the postive number which in this case is 8.6, therefore the equation is 2.3 + 8.6 = 10.9

6 0
3 years ago
Is this correct please give correct answer but no links
natima [27]

Answer:

I give answers I think their correct but as humans we make mistakes so there may be errors

4 0
2 years ago
PLEASE HELP IM DESPERATE precalculus. also 50 points!! please god help
kompoz [17]

Answer:

For B thru F these options will vary but here how you do it

B. Step 1 Draw the 4 Quadrants.

Then Draw the Triangle in the lower right quadrant which we call quadrant 4. Label the X axis as Adjacent and positive. Label the Y axis as Opposite and negative. Label the Slanted side as the hypotune and AS POSITVE SINCE HYPOTENUSE IS ALWAYS POSITIVE.

FOR C. IN QUADRANT 2, PLOT A POINT AT 0,12 AND AT (-5,0). CONNECT THE DOTS AND IT FORMS A TRIANGLE. Label the X axis as adjacent and negative and the y axis as positve and opposite and label the slanted side hypotunese and positive.

FOR D Draw a straight line along the x axis then draw a slanted line passing through (5,-1). In between them put the theta symbol in there.

The labeling is the same for C.

For E. Since tan must be positve and secant must be positve, our triangle must be in the 1st Quadrant. Draw any right triangle as long it is in the first quadrant

The x axis is adjacent and positve. The y axis is opposite and positve. The hypotenuse is the slanted side and it is positve.

For F. Since sin is negative and cos is positve the triangle is in the 4th quadrant. Draw any triangle in the 4th quadrant and the labeling is the same for Problem B.

2. We can find the sec of cos by flipping cosine.

\cos( \frac{x}{y} )  =  \sec( \frac{y}{x} )

\cos( \frac{1}{2} )  =  \sec(2 )

Sec is 2.

To find the cotangent, first let find the sin then tan.

We can use the identity

\cos( {theta}^{2} )  +  \sin( {theta}^{2} )  = 1

Let plug in the number

\cos( \frac{ {1}^{2} }{{2}^{2} } )  +  \sin(x {}^{2} )  = 1

\cos( \frac{1}{4} )  +  \sin(x {}^{2} )  = 1

\sin(x {}^{2} )  =   1 -  \frac{1}{4}

\sin(x {}^{2} )  =  \frac{3}{4}

\sin(x)  =  \frac{ \sqrt{3} }{ \sqrt{4} }

\sin(x)  =  \frac{ \sqrt{3} }{2}

Since sin is negative, sin x=

-  \frac{ \sqrt{3} }{2}

Now let apply the formula

\frac{ \sin(x) }{ \cos(x) }  =  \tan(x)

\frac{ \frac{ -  \sqrt{3} }{2} }{ \frac{1}{2} }  =  \tan(x)

-  \sqrt{3}

Now let find cotangent we can the reciprocal of

tan.

\tan=  -  \sqrt{3}

\cot =   - \frac{1}{ \sqrt{3} }

Rationalize denominator

\frac{ - 1}{ \sqrt{3} }  \times   \frac{ \sqrt{3} }{ \sqrt{3} }  =  \frac{ \sqrt -{ 3} }{3}

cotangent equal

-  \frac{ \sqrt{ 3} }{3}

4 0
3 years ago
How would you solve #2 and #3?
motikmotik
I think u just connect the dots
5 0
3 years ago
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