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Ludmilka [50]
3 years ago
11

Write the equation of a line, in SLOPE INTERCEPT FORM, that is perpendicular to the given equation and passes through the given

point.
Starting equation- y = 1/3 + 4

Given Point- (-5, 2)


Any help with this would be great :)
Mathematics
1 answer:
adoni [48]3 years ago
4 0

Step-by-step explanation:

slope interception formula is

y-y1=m(x-x1)

where m is m=y-y1/x-x1 in this case m=-2 because the line we are trying to find is parallel to the given one y=-2x-6 where slope k=-2

so the final equation would be

y-1=-2(x-(-4))

y-1=-2x-2*4

y=-2x-8+1=-2x-7

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The domain for the following function is ____ ?
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Answer:

All real numbers except x cannot equal 10.

Step-by-step explanation: the denominator can not equal to zero that is undefined.

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Help! I need work too. I have a test tomorrow and I don't know how to solve this.
sleet_krkn [62]
20        7
----- = ------
100      x

100 x 7 = 700
700/20 = 35
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I just need no. a please help me to prove this. ​
OleMash [197]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π    and         cos A = cos B · cos C

scratchwork:

  A + B + C = π

               A = π - (B + C)

         cos A = cos [π - (B + C)]                              Apply cos

                    = - cos (B + C)                                    Simplify

                    = -(cos B · cos C - sin B · sin C)          Sum Identity

                    = sin B · sin C - cos B · cos C               Simplify

cos B · cos C = sin B · sin C - cos B · cos C               Substitution

2cos B · cos C = sin B · sin C                                        Addition

                     2=\dfrac{\sin B\cdot \sin C}{\cos B \cdot \cos C}                                     Division

                     2 = tan B · tan C

\text{Use the Sum Identity:}\quad \tan(B+C)=\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}

<u>Proof LHS → RHS</u>

Given:                              A + B + C = π

Subtraction:                     A = π - (B + C)

Apply tan:                  tan A = tan(π - (B + C))

Simplify:                               = - tan (B + C)

\text{Sum Identity:}\qquad \qquad \qquad =-\bigg(\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}\bigg)

Substitution:                        = -(tan B + tan C)/(1 - 2)

Simplify:                               = -(tan B + tan C)/-1

                                            = tan B + tan C

LHS = RHS:   tan B + tan C = tan B + tan C  \checkmark

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