set f(x) equal to y
y = 19/ x^3
swap x and y
x = 19/y^3
make y the subject
xy^3 = 19
y^3 = 19/x
![y = \sqrt[3]{ \frac{19}{x} }](https://tex.z-dn.net/?f=y%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B19%7D%7Bx%7D%20%7D%20)
then just replace y with f^-1(x)
![f(x) = \sqrt[3]{ \frac{19}{x} }](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B19%7D%7Bx%7D%20%7D%20)
hope this helped! have a good day ~ •lipika•
The answer is 7.68. Since she gave one to her sister, she now has 6 to share with her friends. Now we can conclude the equation to solve this problem would be 6x1.28
Answer:
-3
Step-by-step explanation:
3n-4=-13
+4 +4
3n =-9
3 3
I hope that helped
Answer:
please send full question
Step-by-step explanation:
Answer:
1. 15
2. 8
Step-by-step explanation:
The two sequence are geometric progression GP, because they follow a constant multiple (common ratio)
The nth term of a GP is;
Tn = ar^(n-1)
Where;
a = first term
r = common ratio
For the first sequence;
The common ratio r is
r = T3/T2 = 540/90 = 6
r = 6
T2 = ar^(2-1) = ar
T2 = 90 = ar
Substituting the values of r;
90 = a × 6
a = 90/6
a = 15
First term = 15
2. The sam method applies here.
Common ratio r = T3/T2 = 128/32 = 4
r = 4
T2 = ar^(2-1) = ar
T2 = 32 = ar
Substituting the values of r;
32 = a × 4
a = 32/4
a = 8
First term = 8
