Answer:
y = 2x - 7
Step-by-step explanation:
Parallel lines have the same slope, so the line will also have a slope of 2
Plug in the slope and given point into y = mx + b to solve for b:
y = mx + b
3 = 2(5) + b
3 = 10 + b
-7 = b
Then, plug in the slope and b into y = mx + b
y = 2x - 7 is the equation of the line
Answer:
84 is the highest possible course average
Step-by-step explanation:
Total number of examinations = 5
Average = sum of scores in each examination/total number of examinations
Let the score for the last examination be x.
Average = (66+78+94+83+x)/5 = y
5y = 321+x
x = 5y -321
If y = 6, x = 5×6 -321 =-291.the student cannot score -291
If y = 80, x = 5×80 -321 =79.he can still score higher
If If y = 84, x = 5×84 -321 =99.This would be the highest possible course average after the last examination.
If y= 100
The average cannot be 100 as student cannot score 179(maximum score is 100)
Answer:
CD = 6.385 units
Step-by-step explanation:
Given triangle ABC with right angle at C.
And AB = AD + 6 .
Now, consider the triangle ABC.
⇒ cos(∠BAC) = 
(cosФ = adj/hyp)
cos(20) = .
0.9397 = 
(since AB = AD + 6 and AC = AD + CD)
⇒ 0.9397 AD + 5.6382 = AD + CD
⇒ CD = 0.0603 AD + 5.6382. →→→→→ (1)
⇒ sin(∠BAC) = 
(sinФ = opp/hyp)
sin(20) = .
⇒ BC = AB sin(20) . →→→→→(2)
Now, consider the triangle BCD,
sin(∠BDC) = 
⇒ sin(80) =
CD = 
From (2), CD = 
.
⇒ CD = AB (0.3473)
⇒ CD = (AD + 6) (0.3473)
⇒ CD = 0.3473 AD + 2.0838 →→→→→→(3)
Now, (1) →→ CD = 0.0603 AD + 5.6382
(3) →→ CD = 0.3473 AD + 2.0838
⇒ 0.0603 AD + 5.6382 = 0.3473 AD + 2.0838
0.287 AD = 3.5544.
⇒ AD = 12.3847
⇒ From (1), CD = 0.0603(12.3847) + 5.6382
⇒ CD = 6.385 units
First find the slope between the 2 points
0--2/4--4
=2/8
=1/4
Then y=1/4 x+b
Now substitute one pair of x and y into the equation
0=1/4*4+b
b=-1
So y=1/4x-1
Done!
Answer:
7/10
Step-by-step explanation:
