Answer:
When you're dividing, I only know that one. sorry..
Step-by-step explanation:
Answer:
Step-by-step explanation:
R(0,0)
A=((0+2a)/2,(0+2b)/2)=(a,b)
S(2a,2b)
B=((2a+2c)/2,(2b+2d)/2)=(a+c,b+d)
T(2c,2d)
C=((2c+2c)/2,(2d+0)/2)=(2c,d)
V(2c,0)
D=((2c+0)/2,(0+0)/2)=(c,0)
R(0,0)
slope of AB=(b+d-b)/(a+c-a)=d/c
slope of DC=(d-0)/(2c-c)=d/c
slope of AD=(0-b)/(c-a)=-b/(c-a)
slope of BC=(d-b-d)/(2c-a-c)=-b/(c-a)
Answer:
20.758
Step-by-step explanation:
8.74 + 7.086 + 4.932
(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.
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(b) The following functions are positive in quadrant III:
tangent, cotangent
The following functions are negative in quadrant III
cosine, sine, secant, cosecant
A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.
Answer:
y=2x-11
Step-by-step explanation:
