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:

Step-by-step explanation:
The given equation in rectangular coordinates is;

We use the relation;

and

This implies that;


Divide through by r to get;

Divide both sides by 

Answer:
y=-7
Step-by-step explanation:
hope that will help you
Answer: 0.000
Step-by-step explanation:
Answer:
Option A
Step-by-step explanation:
By applying Pythagoras theorem in ΔABC,
(Hypotenuse)² = (Leg 1)² + (Leg 2)²
AC² = AB² + BC²
AC² = 6² + 5²
AC = 
Similarly, by applying Pythagoras theorem in ΔACD,
AD² = AC² + CD²
(10)² =
+ x²
x² = 100- 61
x² = 39
x = √(39)
Option A will be the answer.