So the slope is 5 and you know the points, (0,4)
y-4=5(x-0)
y-4 = 5x
y = 5x + 4 is the slope intercept form
standard form is: -5x + y = 4 OR y - 5x = 4

5 0

4 years ago

I have a question about inscribed angles...

Vika [28.1K]

Answer:

a=38

b=45

c=38

d=105

e=45

Step-by-step explanation:

5 0

4 years ago

On what intervals is the graph of f(x)=x^3-2x^2+5x+1 increasing?

amid [387]

$f(x)=x^3-2x^2+5x+1\\ f'(x)=3x^2-4x+5\\
3x^2-4x+5=0\\\Delta=(-4)^2-4\cdot3\cdot5=16-60=-44$
$\Delta0 \Rightarrow$ the graph of the parabola is above the x-axis, so the derivative is always positive and therefore the initial function is increasing in its whole domain.

$f(x)=0.5x^2-6\\
f'(x)=x$
The function is decreasing when its first derivative is negative. The first derivative of this function is negative for $x$ so for $x\in(-\infty,0)$ the function is decreasing.

$f(x)=\dfrac{x+1}{x-1}\qquad(x\not=1)\\
f'(x)=\dfrac{x-1-(x+1)}{(x-1)^2}=-\dfrac{2}{(x-1)^2}$
The function is increasing when its first derivative is positive. The first derivative of this function is always negative therefore this function is never increasing.

3 0

4 years ago