Answer:
548 and 213
Step-by-step explanation:
Use a linear function:
Let x = 1st number
Let y = 2nd number
x+y=761
x-y=335
2x=1096
x=548
plug x into either equation
548+y=761
y=761-548
y=213
548-y=335
-y=-213
y=213
If
<em>f(x)</em> = <em>ax</em> ³ + <em>bx</em> ² - 5<em>x</em> + 9
then
<em>f '(x)</em> = 3<em>ax </em>² + 2<em>bx</em> - 5
Given that <em>f</em> (-1) = 12 and <em>f</em> '(-1) = 3, we get the system of equations
-<em>a</em> + <em>b</em> + 5 + 9 = 12
3<em>a</em> - 2<em>b</em> - 5 = 3
or
-<em>a</em> + <em>b</em> = -2
3<em>a</em> - 2<em>b</em> = 8
Multiply through the first equation by 2 and add it to the second one to eliminate <em>b</em> and solve for <em>a</em> :
2(-<em>a</em> + <em>b</em>) + (3<em>a</em> - 2<em>b</em>) = 2(-2) + 8
-2<em>a</em> + 2<em>b</em> + 3<em>a</em> - 2<em>b</em> = -4 + 8
<em>a</em> = 4
Substitute this into the first equation above to solve for <em>b</em> :
-4 + <em>b</em> = -2
<em>b</em> = 2
Answer:
first one
Step-by-step explanation:
Because l//m, s＝n
k＋p＋n＝180
So k＋p＋s＝180
Answer:
x2=−8(y−2)
Step-by-step explanation:
Parabola is a locus of a point which moves at the same distance from a fixed point called the focus and a given line called the directrix.
Let P(x,y) be the moving point on the parabola with
focus at S(h,k)= S(0,0)
& directrix at y= 4
Now,
|PS| = √(x-h)2 + (y-k)2
|PS| = √(x-0)2 + (y-0)2
|PS| = √ x2 + y2
Let ‘d’ be the distance of the moving point P(x,y) to directrix y- 4=0
- d= |ax +by + c|/ √a2 + b2
- d= |y-4|/ √0 + 1
- d= |y-4| units.
equation of parabola is:
- |PS| = d
- √ x2 + y2 = |y-4|
Squaring on both sides, we get:
- x2 + y2 = (y-4)2
- x2 + y2 = y2 -8y + 16
- x2 = - 8y + 16
- x2 = -8 ( y - 2)
This is the required equation of the parabola with focus at (0,0) and directrix at y= 4.
