Answer:
See Below.
Step-by-step explanation:
We are given that ΔAPB and ΔAQC are equilateral triangles.
And we want to prove that PC = BQ.
Since ΔAPB and ΔAQC are equilateral triangles, this means that:
Likewise:
Since they all measure 60°.
Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:
Likewise:
Since ∠QAC ≅ ∠PAB:
And by substitution:
Thus:
Then by SAS Congruence:
And by CPCTC:
Answer:
Angle A is 101
Angle B is 79
Angle C is 83
Angle D is 97
Step-by-step explanation:
Vertical angles are congruent, which applies to C.
Lines are parallel so 79 and angle B are the same
Angle a is 101 because its a supplementary angle
Angle D is 97 because its also a supplementary angle so you subtract 83 from 180.
Answer:
The answer to your question is a) y - 2 = -5(x - 3)
b) y = -5x + 17
Step-by-step explanation:
Data
A (1, 12)
B (3, 2)
Process
1.- Find the slope
Formula
Substitution
2.- Find the equation in point slope form
Equation
y - y1 = m(x - x1)
Substitution and equation
y - 2 = -5(x - 3)
3.- Find the slope-intercept form
Expand the point slope form
y - 2 = -5x + 15
Simplify
y = -5x + 15 + 2
Equation
y = -5x + 17
A cosine is just a sine shifted to the left by π/2. A cosine of 4x is shifted to the left by only π/8 because of the factor 4. Sketch them.
The region we're looking for is this sausage-shaped part between the cos and the sin.
The x intercepts are at π/8 for the cosine and π/4 for the sine. The midpoint between them is at (π/8 + π/4)/2 = 3/16π.
The region is point symmetric around the x axis, so the y coordinate of the centroid is 0.
So the centroid is at (3/16π, 0)
Answer:grater than 8 less than 9
Step-by-step explanation:
had the same question
