![\dfrac{\sqrt[3]{x+h}-\sqrt[3]x}h\times\dfrac{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}=\dfrac{(\sqrt[3]{x+h})^3-(\sqrt[3]x)^3}\cdots](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7Bx%2Bh%7D-%5Csqrt%5B3%5Dx%7Dh%5Ctimes%5Cdfrac%7B%5Csqrt%5B3%5D%7B%28x%2Bh%29%5E2%7D%2B%5Csqrt%5B3%5D%7Bx%28x%2Bh%29%7D%2B%5Csqrt%5B3%5D%7Bx%5E2%7D%7D%7B%5Csqrt%5B3%5D%7B%28x%2Bh%29%5E2%7D%2B%5Csqrt%5B3%5D%7Bx%28x%2Bh%29%7D%2B%5Csqrt%5B3%5D%7Bx%5E2%7D%7D%3D%5Cdfrac%7B%28%5Csqrt%5B3%5D%7Bx%2Bh%7D%29%5E3-%28%5Csqrt%5B3%5Dx%29%5E3%7D%5Ccdots)

The

s then cancel, leaving you with the
![\cdots=\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Ccdots%3D%5Csqrt%5B3%5D%7B%28x%2Bh%29%5E2%7D%2B%5Csqrt%5B3%5D%7Bx%28x%2Bh%29%7D%2B%5Csqrt%5B3%5D%7Bx%5E2%7D)
term.
If it's not clear what I did above, consider the substitution
![a=\sqrt[3]{x+h}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7Bx%2Bh%7D)
and
![b=\sqrt[3]x](https://tex.z-dn.net/?f=b%3D%5Csqrt%5B3%5Dx)
. Then

We have that
csc ∅=13/12
sec ∅=-13/5
cot ∅=-5/12
we know that
csc ∅=1/sin ∅
sin ∅=1/ csc ∅------> sin ∅=12/13
sec ∅=-13/5
sec ∅=1/cos ∅
cos ∅=1/sec ∅------> cos ∅=-5/13
sin ∅ is positive and cos ∅ is negative
so
∅ belong to the II quadrant
therefore
<span>the coordinates of point (x,y) on the terminal ray of angle theta are
</span>x=-5
y=12
the answer ispoint (-5,12)
see the attached figure
Answer:
is an asymptote
Step-by-step explanation:
Apply a table, and plug in values of x into the equation. For example f(2)=
. That's one of the points on the graph that we just found (2,6)
Answer:
5.
1. a
2.d
3.b
4.e
5.c
Step-by-step explanation:
can't do question 6 though, sorry about that