Answer:
The answer should be C. By combining like terms you can find the answer. 1.75a+2.75a = 4.5a
2.25b + 1.75b = 4b
2.25c + 1.25c = 3.5c
Step-by-step explanation:
Answer:
Option (1) will be the answer.
Step-by-step explanation:
Coordinates of the points A and B lying on the line f are (0, 2) and (2, 0) respectively.
Slope of the line f,


After dilation of line f by a scale factor of 2, coordinates of A' and B' will be,
Rule for dilation,
(x, y) → (kx, ky)
Where k = scale factor
A(0, 2) → A'(0, 4)
B(2, 0) → B'(4, 0)
Slope of line f',


Since, 
Therefore, both the lines f and f' will be parallel.
Option (1) will be the answer.
Answer:
450,000 codes
Step-by-step explanation:
Number of digits in the access code = 6
Conditions:
- First digit cannot be 2
- Last digit must be odd
Since, there are 10 digits in total from 0 to 9 and first digit cannot be 2, there are 9 ways to fill the place of first digit.
Last digit must be odd. As there are 5 odd digits from 0 to 9, there are 5 ways to fill the last digit.
The central 4 digits can be filled by any of the 10 numbers. So, each of them can be filled in 10 ways.
According to the fundamental rule of counting, the total possible codes would be the product of all the possibilities of individual digits.
Therefore,
Number of possible codes = 9 x 10 x 10 x 10 x 10 x 5 = 450,000 codes
Hence, 450,000 different codes are possible for the lock box
Answer:
Step-by-step explanation:
So, after you roll it, the width will be the circumference of the circle 
The minimum amount will be 
I don't know, how number you want to round, so you can round it by yourself
Answer:
18 years ( approx )
Step-by-step explanation:
Since, if an amount is compounded daily,
Then, the final amount after t years is,
A= P(1 + r/365) 365t
Where, P is the principal amount
r is the annual rate ( in decimals ),
t is the number of years,
Given,
A = $ 800,
r = 3.9% = 0.039,
P = $ 400,
By substituting the values,
800= 400 (1 + 0.039/365)365t
t = 17.774 = 18
Hence, his balance will be $800 after approximately 18 years.