Let F(x)=x^2-13 and G(x)=18-x

Find (F⋅G)(-1)=

If f(2) = 5, f(3) = 2, f(4) = 5, g(2) = 6, g(3) = 2 and g(4) = 0, find (f · g)(2) + f(g(3)). ... f(x) = { c2x − c x ≤ 1 cx − x x > 1. Solution. The partial functions of f(x) are continuous for x < 1 and x > 1 ... 5 , we have f (x) = 5. 3 x. 2. 3 +. 6(−3. 5. )x−8. 5 = 5. 3√ x2. 3. − 18. 5. 5√ x8 . 8. ... x2 − 2x = −1 =⇒ (x − 1)(x − 1)=0=⇒ x = 1.

Step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

believe it or not anna is correct good job anna give me a fist bump there u go gal