Answer:
bottom of graph will move from (0,0) to point (1,3) after transformation
Step-by-step explanation:
given
original : f(x) = 
transformed; g(x) = 
+ 3
look at this way g(x) = 
+ k
if (x-h), h>0, move h units to the right
if k>0, move k units up
the bottom of the graph will be at point (1,3)
Let's begin by listing the first few multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 38, 40, 44. So, between 1 and 37 there are 9 such multiples: {4, 8, 12, 16, 20, 24, 28, 32, 36}. Note that 4 divided into 36 is 9.
Let's experiment by modifying the given problem a bit, for the purpose of discovering any pattern that may exist:
<span>How many multiples of 4 are there in {n; 37< n <101}? We could list and then count them: {40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100}; there are 16 such multiples in that particular interval. Try subtracting 40 from 100; we get 60. Dividing 60 by 4, we get 15, which is 1 less than 16. So it seems that if we subtract 40 from 1000 and divide the result by 4, and then add 1, we get the number of multiples of 4 between 37 and 1001:
1000
-40
-------
960
Dividing this by 4, we get 240. Adding 1, we get 241.
Finally, subtract 9 from 241: We get 232.
There are 232 multiples of 4 between 37 and 1001.
Can you think of a more straightforward method of determining this number? </span>
Step-by-step explanation:
radius(CE)=diameter/2=9
EF=8inch
DE=8/2=4inch
pythagorean theorem:
im right triangle,
a^2+b^2=c^2
4^2+b^2=9^2
16+b^2=81
b^2=81-16
b^2=65
b=√65
√65≈8.06
