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lord [1]
3 years ago
11

point f has coordinated (8,-1), point g is symmetric to point f with respect to the line y=x, what are the coordinates of point

g
Mathematics
1 answer:
Alex17521 [72]3 years ago
7 0
If points f and g are symmetric with respect to the line y=x, then the line connecting f and g is perpendicular to y=x, and f and g are equidistant from y=x.

This problem could be solved graphically by graphing y=x and (8,-1).  With a ruler, measure the perpendicular distance from y=x of (8,-1), and then plot point g that distance from y=x in the opposite direction.  Read the coordinates of point g from the graph.

Alternatively, calculate the distance from y=x of (8,-1).  As before, this distance is perpendicular to y=x and is measured along the line y= -x + b, where b is the vertical intercept of this line.  What is b?  y = -x + b must be satisfied by (8,-1):  -1 = -8 + b, or b = 7.  Then the line thru (8,-1) perpendicular to y=x is     y = -x + 7.  Where does this line intersect y = x?

y = x = y = -x + 7, or 2x = 7, or x = 3.5.  Since y=x, the point of intersection of y=x and y= -x + 7 is (3.5, 3.5).

Use the distance formula to determine the distance between (3.5, 3.5) and (8, -1).  This produces the answer to this question.

 
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Expand the left and right sides.
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Rudiy27

Answer:

Her usual driving speed is 38 miles per hour.

Step-by-step explanation:

We know that:

s = \frac{d}{t}

In which s is the speed, in miles per hour, d is the distance, in miles, and t is the time, in hours.

We have that:

At speed s, she takes two hours to drive. So

s = \frac{d}{2}

d = 2s

However, on one particular trip, after 40% of the drive, she had to reduce her speed by 30 miles per hour, driving at this slower speed for the rest of the trip. This particular trip took her 228 minutes.

228 minutes is 3.8 hours. So

0.4s + 0.6(s - 30) = \frac{d}{3.8}

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Her usual driving speed is 38 miles per hour.

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Step-by-step explanation: do you have any choices


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If tanA=a <br>then find sin4A-2sin2A/ sin4A+2sin2A​
anygoal [31]

Answer:

The value of the given expression is

\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

Step by step Explanation:

Given that tanA=a

To find the value of \frac{sin4A-2sin2A}{sin4A+2sin2A}

Let us find the value of the expression :

\frac{sin4A-2sin2A}{sin4A+2sin2A}=\frac{2cos2Asin2A-2sin2A}{2cos2Asin2A+2sin2A} ( by using the formula sin2A=2cosAsinA here A=2A)  

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=\frac{(-(1-cos2A))}{(1+cos2A)}(using  sin^2A+cos^2A=1  here A=2A)

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=\frac{-(sin^2A+cos^2A-cos^2A+sin^2A)}{sin^2A+cos^2A+(cos^2A-sin^2A)}

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=\frac{-2sin^2A}{2cos^2A}

=-\frac{sin^2A}{cos^2A}

=-tan^2A  ( using tanA=\frac{sinA}{cosA} here A=2A )

 =-a^2 (since tanA=a given )

Therefore \frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

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