A rhombus has four equal sides. If the perimeter of this rhombus is 164, then the length of one side is 164/4, or 41.
Draw this rhombus. Label all four sides with "41." Label the longer diagonal 80 and the half length of that diagonal 40. You will see inside the rhombus four congruent triangles with hypotenuse 41, leg 10 and unknown height. Thus, this unknown height is found by solving x^2 + 40^2 = 41^2, and x^2=9, so that the length of the shorter diagonal is 2(2) = 18 (answer).
There are 720 total different combinations, and 24 of those will have the letters DAY consecutively within all those combinations.
<u>Explanation:</u>
FRIDAY is a six letter word.
To form all combinations of letters, we would have six choices of letters to fill the first spot. Whatever letter you choose for the first spot, you then have five choices of letters to fill the second spot. Four choices of letters to fill the third spot. And so on.
So, the total number of different combinations of the six letters would be 6!
6! = 6 X 5 X 4 X 3 X 2 X 1
= 720
DAY could appear consecutively starting in 4 different spots: the 1st, 2nd, 3rd, and 4th spots of all combinations. DAY will appear consecutively 6 times in each of those 4 starting spots.
6 X 4 = 24.
So, there are 720 total different combinations, and 24 of those will have the letters DAY consecutively within all those combinations.
Answer:
Please refer to the attachment
The correct answer is C because when using distributive property it simplifies back to 5b+30