(2) + (2)
=> 8x - 6y = -46 ———(3)
(1) + (3)
=> 5x + 6y + (8x - 6y) = 20 - 46
=> 13x + (6-6)y = -26
=> 13x + 0y = -26
=> 13x = -26
=> x = -26/13
=> x = -2
=> 5(-2) + 6y = 20
=> 6y = 20 + 10
=> 6y = 30
=> y = 5
x = -2, y = 5 => C
We are given the function <span>f(x)=sqrt of (4sinx+2) and is asked to find the first derivative of the function when x is equal to zero.
</span><span>f(x)=sqrt of (4sinx+2)
f'(x) = 0.5 </span><span>(4sinx+2) ^ -0.5 * (4cosx)
</span>f'(0) = 0.5 <span>(4sin0+2) ^ -0.5 * (4cos0)
</span>f'(0) = 0.5 <span>(0+2) ^ -0.5 * (4*1)
</span>f'(x) = 0.5 (2) ^ -0.5 * (4)
f'(x) = -.1.65
13) range: (−2,∞)
x-intercept: (1n(2),0)
y-intercept: (0,-1)
asymptote: y = -2
14) range: (−∞,2)
x-intercept: (1n(2/3),0)
y-intercept: (0,-1)
asymptote: y = 2
I got these answers on M a t h w a y. It is really helpful for these types of problems.
Hope this helps✨❤️
Easy peasy
the bit where it says
D={something}
those are the numbers you should input for x to get y values
3.
first solve for y to make life easier
-3x-5y=20
times -1
3x+5y=-20
minus 3x both sides
5y=-3x-20
divide both sides by 5
y=-3/5x-4
sub value of the domain
x=-10
y=-3/5(-10)-4
y=6-4
y=2
a point is (-10,2)
x=-5
y=-3/5(-5)-4
y=3-4
y=-1
(-5,-1) is another
x=0
y=-3/5(0)-4
y=-4
(0,-4) is another
x=5
y=-3/5(5)-4
y=-3-4
y=-7
(5,-7)
points are (-10,2), (-5,-1), (0,-4), (5,-7)
4.
input
x=-2
y=(-2)^2-3
y=4-3
y=1
(-2,1)
x=-1
y=(-1)^2-3
y=1-3
y=-2
(-1,-2)
x=0
y=(0)^2-3
y=-3
(0,-3)
x=1
y=(1)^2-3
y=1-3
y=-3
(1,-3)
x=2
y=(2)^2-3
y=4-3
y=1
(2,1)
the points are (-2,1), (-1,-2), (0,-3), (1,-2), (-2,1)
see graph below
Answer:
slope of perpendicular line - m = -1/2
passing through (-2,3)
thus,
y-3 = -1/2 (x+2)
or y = -1/2x +2