Which graph represents the inequality? y≤13x−1
2 answers:
Answer:
The graph in the attached figure
Step-by-step explanation:
we have
we know that
The solution of the inequality is the shaded area below the solid line
The slope of the line is positive
The equation of the line is equal to
The y-intercept of the line is the point
The x-intercept of the line is the point
therefore
The graph in the attached figure
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Answer:
see below
Step-by-step explanation:
y = 8x − 9
y = 4x − 1
Since both equations are equal to y, we can set them equal to each other
8x-9 = 4x-1
Subtract 4x from each side
8x-4x-9 = 4x-4x-1
4x-9 = -1
Add 9 to each side
4x-9+9 = -1+9
4x = 8
Divide each side by 4
4x/4 = 8/4
x=2
Now find y
y = 4x-1
y = 4(2)-1
y = 8-1
y = 7
(2,7)
The point of intersection when the 2 lines are graphed is (2,7)
Domain is the numbers yo can use
we has a sqrt and a ln
we cant have any neagtive ln's or sqrts
x cannot be negative
it cannot be 0 either
domain is all numbers bigger than 0
deritivive
apply chain rule
dy/dx(f(g(x))=f'(g(x))g'(x)
and also
dy/dx (lnx)=1/x
and
dy/dx(f(x)g(x))=f'(x)g(x)+g'(x)f(x)
and from experience
dy/dx(sqrtx)=1/(2√x)
so
dy/dx(ln(2xsqrt(2+x)))=
(1/(2xsqrt(2+x))(2)(sqrt(2+x)+(x/(2sqrt(2+x))=
(3x+4)/(2x(x+2))
domain is all real numbers greater than 0
derivitive is
we has a sqrt and a ln
we cant have any neagtive ln's or sqrts
x cannot be negative
it cannot be 0 either
domain is all numbers bigger than 0
deritivive
apply chain rule
dy/dx(f(g(x))=f'(g(x))g'(x)
and also
dy/dx (lnx)=1/x
and
dy/dx(f(x)g(x))=f'(x)g(x)+g'(x)f(x)
and from experience
dy/dx(sqrtx)=1/(2√x)
so
dy/dx(ln(2xsqrt(2+x)))=
(1/(2xsqrt(2+x))(2)(sqrt(2+x)+(x/(2sqrt(2+x))=
(3x+4)/(2x(x+2))
domain is all real numbers greater than 0
derivitive is