The direction vector of the line
L: x=1+t, y=4t, z=2-3t
is <1,4,-3>
which is also the required normal vector of the plane.
Since the plane passes through point (-5,9,10), the required plane is :
Π 1(x-(-5)+4(y-9)-3(z-10)=0
=>
Π x+4y-3z=1
Answer:
Slope(m) = 16
Y intercept (c) = 23
Step-by-step explanation:
Given:
Points : (x1, y1) = (2, 55)
(x2, y2) = (4, 87)
Using two point form to calculate slope:
m = 16
To find y intercept:
Take any one point ans slope :
Using slope point formula:
y = mx + c
87 = 16(4) + c
Solving for c:
c = 23
Answer: y = -2x + 16
Explanation:
Use point slope form:
(y1 - y) = m(x1 - x)
y1 = 6 and x1 = 5 and m = -2
6 - y = -2(5 - x)
6 - y = -10 + 2x
-y = 2x - 10 - 6
-y = 2x - 16
y = -2x + 16
Answer:
see explanation
Step-by-step explanation:
Using the trigonometric identity
sin²x + cos²x = 1
Consider the left side
2cos²A - 1 ← replace 1 by sin²A + cos²A
= 2cos²A - ( sin²A + cos²A)
= 2cos²A - sin²A - cos²A
= cos²A - sin²A = right side ⇒ verified
