Answer:
hey i hope this helps u !!!
Step-by-step explanation:
An integer may be a multiple of 3.
An integer may be 1 greater than a multiple of 3.
An integer may be 2 greater than a multiple of 3.
It is redundant to say an integer is 3 greater than a multiple of 3 (that's just a multiple of 3, we've got it covered). Same for 4, 5, 6, 7...
Let's consider a number which is a multiple of 3. Clearly, we can write 3+3+3+3+... until we reach the number. It can be written as only 3's.
Let's consider a number which is 2 greater than a multiple of 3. If we subtract 5 from that number, it'll be a multiple of 3. That means we can write the number as 5+3+3+3+3+... Of course, the number must be at least 8.
Let's consider a number which is 1 greater than a multiple of 3. If we subtract 5 from that number, it'll be 2 greater than a multiple of 3. If we subtract another 5, it'll be a multiple of 3. That means we can write the number as 5+5+3+3+3+3+... Of course, the number must be at least 13.
That's it. We considered all the numbers. We forgot 9, 10, 11, and 12, but these are easy peasy.
Beautiful question.
Answer:
I = $1,200.00
Step-by-step explanation:
First, converting R percent to r a decimal
r = R/100 = 6%/100 = 0.06 per year,
then, solving our equation
I = 4000 × 0.06 × 5 = 1200
I = $1,200.00
Answer:
16
Step-by-step explanation:
We can see that Ar is the perpendicular bisector of chord BD. Since A is the center of the circle, AR is the radius of the circle, which is 10 (6+4)
Next, we can see that when we connect point A to point D, it is also a radius. Thus, AD is also equal to 10 as the radius of the circle remains the same.
Using Pythagoras theorem, a^2 + b^2 = c^2, we can make a right angled triangle of ACD.
AC = 6 = a
CD = ? = b
AD = 10 = c
10^2 = 6^2 + b^2
b^2 = 10^2 - 6^2 = 64
b = CD = 8
Now, since Ar is the perpendicular bisector of chord BD, BD = CD x 2
BD = 8 x 2 = <u>16</u>