Can anyone else figure this out
Answer:
Please find attached, the required drawing of quadrilateral ABCD and the dilation of quadrilateral ABCD scaled down by a scale factor of 1/3
Step-by-step explanation:
The coordinates of the quadrilateral ABCD are;
A(0, 6), B(6, 6), C(9, 0), D(0, 0)
From the dilation by a scale factor of 1/3, using D as the center of dilation, we have from the attached drawing, the following coordinates of the quadrilateral EFGH as E(0, 2), F(2, 2), G(3, 0) H(0, 0) which is the quadrilateral ABCD scaled down by 1/3
Answer: Please see explanation column for answer
Step-by-step explanation:
According to sequence 2, 6, 18, 54, 162
--we can see that it is a geometric sequence since to go from one term to the next requires us to multiply by 3 which is called the common ratio which is gotten by dividing the next by the previous.
ie term 2/ term 1= 6/2 = 3
term 3/ term 2= 18/6= 3
term 4/ term 3= 54/28= 3
term 5 / term 4= 162/54=3
term 6 will therefore be 162 x3 = 486
and term 7 = 486 x 3=1,458 and so on
Recall that nth term of GP is
Tn = ar^n-1
Where a= first term
r= common ratio
n= term
such that to easily find particular term, we plug in the values and calculate
for example the 10th term of the above sequence
Tn = ar^n-1
= 2 X 3 ^(10-1)
2 X 3^9
T10 =39,366
6 cause all the others are 2 digit (i think this is the right one) tell me if i'm wrong
