Answer:
<h2>A</h2>
Step-by-step explanation:
![f(x)=5(3)^x-1\\\\\text{Put x = 0 to the equation of the function:}\\\\f(0)=5(3)^0-1=5(1)-1=5-1=4\to(0,\ 4)\\\\\text{The curve on graph A goes through point (0, 4).}](https://tex.z-dn.net/?f=f%28x%29%3D5%283%29%5Ex-1%5C%5C%5C%5C%5Ctext%7BPut%20x%20%3D%200%20to%20the%20equation%20of%20the%20function%3A%7D%5C%5C%5C%5Cf%280%29%3D5%283%29%5E0-1%3D5%281%29-1%3D5-1%3D4%5Cto%280%2C%5C%204%29%5C%5C%5C%5C%5Ctext%7BThe%20curve%20on%20graph%20A%20goes%20through%20point%20%280%2C%204%29.%7D)
Answer:
![(x + 2)(x + 1)](https://tex.z-dn.net/?f=%28x%20%2B%202%29%28x%20%2B%201%29)
Step-by-step explanation:
by factorising ;
=》
![{x}^{2} + 3x + 2](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%2B%203x%20%20%2B%202)
=》
![{x}^{2} + 2x + x + 2](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%2B%202x%20%2B%20x%20%2B%202)
=》
![x(x + 2) + 1(x + 2)](https://tex.z-dn.net/?f=x%28x%20%2B%202%29%20%2B%201%28x%20%2B%202%29)
=》
![(x + 2)(x + 1)](https://tex.z-dn.net/?f=%28x%20%2B%202%29%28x%20%2B%201%29)
<em>i</em><em> </em><em>hope</em><em> </em><em>it</em><em> </em><em>helped</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer:
tanx·secx
Step-by-step explanation:
To simply this, you can begin by factoring sin x out of the numerator to become:
Now, using Pythagorean Trig Identities, we know that sin²x+cos²x equals 1. We can substitute this to make the equation become:
![\frac{sinx}{cos^{2}x }](https://tex.z-dn.net/?f=%5Cfrac%7Bsinx%7D%7Bcos%5E%7B2%7Dx%20%7D)
First of all, we can convert
to tanx. However, we have a remaining
which, using reciprocal identities, will become sec x.
Finally, we get our answer as tanx·secx.