in order to solve these equations, you need to divide the number by the varible. so #5 is 2.5y=5. you would divide 5 by 2.5, to get 2. the answer would be y=2

8 0

3 years ago

How do I find the scale of the dilation A(3,4),A’(9,12)

Pavlova-9 [17]

Since A (3,4) A’(9,12)
You can see 3*3=9
Therefor it was enlarged by 3times
Scale factor of 3

3 0

3 years ago

Given the tables for f & g below, find the following: x f ( x ) g ( x ) 0 9 7 1 4 8 2 3 6 3 5 1 4 7 5 5 1 2 6 6 0 7 0 4 8 2

d1i1m1o1n [39]

$\\(i)\\
\text{We know that the average rate of change of a function f(x) on inerval [a,b] is}\\
\\
f_{avg}=\frac{f(b)-f(a)}{b-a}\\
\\
\text{so using this the average rate of chang of g for x on interval }[3,7] \text{ is}\\
\\
g_{avg}=\frac{g(7)-g(3)}{7-3}\\
\\
=\frac{4-1}{4}\\
\\
=\frac{3}{4}\\
\\
\text{so average rate of change of g on interval [3,7] is }\frac{3}{4}$

$\\(ii)\\ \text{Average rate of chang of f for x on interval }[3,7] \text{ is}\\
\\
f_{avg}=\frac{f(7)-f(3)}{7-3}\\
\\
=\frac{0-5}{4}\\
\\
=\frac{-5}{4}\\
\\
\text{so average rate of change of g on interval [3,7] is }\frac{-5}{4}$

$\\(iii)\\
\text{Since g(4)=5, so we have}\\
\\
f(g(4))=f(5)\\
\\
\text{since the value of f is 1 at x=5. so }\\
\\
f(g(4))=f(5)=1\\
\\
\text{hence we have, }f(g(4))=$ 1

$\\(iv)\\
\text{Since f(3)=5, so we have}\\
\\
g(f(3))=g(5)\\
\\
\text{since the value of g is 2 at x=5. so }\\
\\
g(f(3))=g(5)=2\\
\\
\text{hence we have, }g(f(3))=$ 2

5 0

3 years ago