77/15 as a mixed number
2 answers:
Answer:
5
Step-by-step explanation:
A mixed number consists of a <u>whole number</u> and a <u>proper fraction</u> (meaning the numerator is less than the denominator).
To get a mixed number, we are going to divide 77/15 using long division:
(I attached an image down below for you to understand better)
15 goes into 77 only 5 times with a remainder of 2.
So our whole number is 5
And the remainder of 2 become of the numerator of our proper fraction, with 15 being our denominator.
So your answer is 5
Hope it helps (●'◡'●)
You might be interested in
Answer:
B.
Step-by-step explanation:
The given circle is centered at and has radius 3 units.
The equation is given by
where (a,b) is the center and r is the radius.
We substitute the values to obtain;
Check the picture below.
segments AC and BC are equal, that means that ABC is an isosceles triangle, with twin sides. Now the twin sides make twin angles on the opposite side, namely the angles at A and B are twins.
if the angles at A and B are twins, and CD is ⟂ to AB, there's only one possibility that can happen, and is if CD is an angle bisector at C.
an angle bisector like CD with twin angles on each side, will cut AB in two equal halves, therefore, if AB = 4, then AD = 2 and DB = 2.
segments AC and BC are equal, that means that ABC is an isosceles triangle, with twin sides. Now the twin sides make twin angles on the opposite side, namely the angles at A and B are twins.
if the angles at A and B are twins, and CD is ⟂ to AB, there's only one possibility that can happen, and is if CD is an angle bisector at C.
an angle bisector like CD with twin angles on each side, will cut AB in two equal halves, therefore, if AB = 4, then AD = 2 and DB = 2.
The first digit can only range from 1-9, resulting in 9 possible options. The next four digits can range from 0-9, resulting in 10 options for each. Since the last two digits remain the same, they do not affect the sample size. Using the fundamental counting principle, we can find the amount of telephone numbers possible in the following equation.
