1.
Let N be the number of coins of Alex
2a+1=N
3b+1=N
4c+1=N
5d+1=N
6e+1=N
7f+1=N
so N-1 is a multiple of 2, of 3, of 4, of 5, of 6 and of 7
so N-1 has factors 7*6*5*2*X, where X is another factor.
2.
There is no need to write N-1 as 7*6*5*3*2*4, since being a multiple of 6, means being a multiple of 2, and of 3 as well. Also, no need to write *4, since 6 has already *2, but we need one more 2, that why we write 7*6*5*2*X
(7*6*5*2=420 is the LCM, least common multiple of 2,3,4,5,6,7)
3.
N-1=420*X.
X cannot be 2 or more since N would become larger that 800.
X must be 1, so N-1=420, which means N=421 is the number of coins
One solution: Two angles of the triangle can be found, since we know their supplement angle counterparts. The first one is 50 degrees, and the second angle is 46 degrees. Thus the final angle has a measurement of 180 - 46 - 50 = 84 degrees. However, "x" is the supplement of the third angle, so 180 - 84 = 96
"x" can also be the sum of the two other angles within the triangle, or 360 - 134 -130, cause all interior angles of any polygon always add up to 360.
Answer:
the quotient of Three-fourths and StartFraction 5 Over 6 EndFract is ![\frac{9}{10}](https://tex.z-dn.net/?f=%5Cfrac%7B9%7D%7B10%7D)
Hence Option B is correct.
Step-by-step explanation:
We need to find quotient of Three-fourths and StartFraction 5 Over 6 EndFract
Writing in mathematical form: ![\frac{3}{4} \div \frac{5}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B4%7D%20%5Cdiv%20%5Cfrac%7B5%7D%7B6%7D)
Solving, we get
(Converting division sign into multiplication the term is reciprocated so, 5/6 will become 6/5 )
![\frac{3}{4} \div \frac{5}{6}\\=\frac{3}{4} \times \frac{6}{5}\\=\frac{3}{2} \times \frac{3}{5}\\=\frac{9}{10}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B4%7D%20%5Cdiv%20%5Cfrac%7B5%7D%7B6%7D%5C%5C%3D%5Cfrac%7B3%7D%7B4%7D%20%5Ctimes%20%5Cfrac%7B6%7D%7B5%7D%5C%5C%3D%5Cfrac%7B3%7D%7B2%7D%20%5Ctimes%20%5Cfrac%7B3%7D%7B5%7D%5C%5C%3D%5Cfrac%7B9%7D%7B10%7D)
So, the quotient of Three-fourths and StartFraction 5 Over 6 EndFract is ![\frac{9}{10}](https://tex.z-dn.net/?f=%5Cfrac%7B9%7D%7B10%7D)
Hence Option B is correct.
Pretty sure it’s C. Quadratic