Answer: The Answers Are: A, C, E
Step-by-step explanation: Find the Equivalent Ratios! :)
Answer:
Step-by-step explanation:
From the graph attached,
Coordinates of the vertices are,
Q(1, 3), R(3, -3), S(0, -2) and T(-2, 1)
Following the rule of translation by 3 units to the right and 2 units down 
(x, y) → (x+3, y-2)
Q(1, 3) → Q''(4, 1)
R(3, -3) → R"(6, -5)
S(0, -2) → S"(3, -4)
T(-2, 1) → T"(1, -1)
Following rule 
(rotation of a point by 180° about the origin) will give the image points,
(x, y) → (-x, -y)
Q"(4, 1) → Q'(-4, -1)
R"(6, -5) → R'(-6, 5)
S"(3, -4) → S'(-3, 4)
T"(1, -1) → T'(-1, 1)
This would be point-slope form.
Slope-intercept form is in the form: y = mx + b
Standard is in the form ax + by = c
Point-slope form is in the form y - y.1 =m(x - x.1)
The form this most closely resembles is point-slope.
Hello from MrBillDoesMath!
Answer:
See Discussion below
Discussion:
(sinq + cosq)^2 = => (a +b)^2 = a^2 + 2ab + b^2
(sinq)^2 + (cosq)^2 + 2 sinq* cosq => as (sinx)^2 + (cosx)^2 = 1
1 + 2 sinq*cosq (*)
Setting a = b = q in the trig identity:
sin(a+b) = sina*cosb + cosa*sinb
sin(2q) = (**)
sinq*cosq + cosq*sinq => as both terms are identical
2 sinq*cosq
Combining (*) and (**)
(sinq + cosq)^2 = 1 + 2sinq*cosq => (**) 2sinq*cosq = sqin(2q)
= 1 + sin(2q)
Hence
(sinq + cosq)^2 = 1 + sin(2q) => subtracting 1 from both sides
(sinq + cosq)^2 - 1 = sin(2q)
The last statement is what we are trying to prove.
Thank you,
MrB
