Answer:
9a) 6 miles
9b) 12:20
Step-by-step explanation:
for first question, use the distance formula: distance = rate · time
d = 2 x 3
d = 6 miles
for second question, she began walking home at 11:05 and it took 1 hour, 15 minutes (15 minutes is 1/4 of 60 minutes)
11:05 + 1 hour = 12:05
12:05 + 15 minutes = 12:20
Answer:
d
Step-by-step explanation:
The rule (x, y ) → (x + 4, y - 6 )
means add 4 to the original x- coordinate and subtract 6 from the original y- coordinate, thus
P(- 8, 3 ) → P'(- 8 + 4, 3 - 6 ) → P'(- 4, - 3 )
Q(- 8, 6 ) → Q'(- 8 + 4, 6 - 6 ) → Q'(- 4, 0 )
R(- 3, 6 ) → R'(- 3 + 4, 6 - 6 ) → R'(1, 0 )
So it's really easy to do this
4 times 2 is 12 and then time 10 degree
=12 times 10 degree
=120 degree
As long any understand that you don't have to change or do anything with the degree and just multiply the numbers .
I hope it helped u
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B = C → A = C - B
→ B = C - A
Use the Double Angle Identity: cos 2A = 2 cos² A - 1
→ (cos 2A + 1)/2 = cos² A
Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]
Use Even/Odd Identity: cos (-A) = cos (A)
<u>Proof LHS → RHS:</u>
LHS: cos² A + cos² B + cos² C
![\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5Cquad%201%2B%5Cdfrac%7B1%7D%7B2%7D%5Cbigg%5B2%5Ccos%20%5Cbigg%28%5Cdfrac%7B2A%2B2B%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B2A-2B%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%2B%5Ccos%5E2%20C%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D1%2B%5Ccos%20%28A%2BB%29%5Ccdot%20%5Ccos%20%28A-B%29%2B%5Ccos%5E2%20C)
![\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cqquad%20%5Cqquad%201%2B%5Ccos%20C%5B%5Ccos%20%28A-B%29%2B%5Ccos%20C%5D)
![\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5Cquad%201%2B%5Ccos%20C%5Cbigg%5B2%5Ccos%20%5Cbigg%28%5Cdfrac%7BA-B%2BC%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BA-B-C%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D1%2B2%5Ccos%20C%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2B%28C-B%29%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B-B-%28C-A%29%7D%7B2%7D%5Cbigg%29)
LHS = RHS: 1 + 2 cos A · cos B · cos C = 1 + 2 cos A · cos B · cos C 
(I'm learning this right now!)
y = 3x - 3
If a line is perpendicular to another line, they have negative reciprocal slopes.
So you would use the slope -1/3.
Use y - y₁ = m(x - x₁) to get an equation in slope intercept form.
Plug the numbers in.
y - 1 = -1/3(x - 3)
y - 1 = -1/3x + 1
y = -1/3x + 2
An equation of a line perpendicular to line CD is y = -1/3x + 2.
