What's eight ninths plus eleven twelfth plus five sixths

What's eight ninths plus eleven twelfth plus five sixths

17/9
8/9+11/12+5/6=96/108+99/108+90/108=285/108
If the mathematical problem is known as X & # 178; = 7.065, how much is x equal to
The result is kept to one decimal place
x²＝7.065
x=±2.7
What is one sixth minus four ninth plus five sixth minus one ninth?
1 / 6 minus 4 / 9 plus 5 / 6 minus 1 / 9 equals 4 / 9
1/6-4/9+5/6-1/9
=1/6+5/6-(4/9+1/9)
=1-5/9=4/9
If Sina = 2cosa, 2 / 3sin ^ 2A + 1 / 4cos ^ 2a, then the value of is
sina=2cosa
Because Sin & sup2; a + cos & sup2; a = 1
So 4cos & sup2; a + cos & sup2; a = 1
cos²a=1/5
sin²a=1-1/5=4/5
So the original formula = 8 / 15 + 1 / 20 = 7 / 12
There are 49 students in class 6 (1). Boys account for 4 / 7 of the total number, and girls account for a few percent of boys
Boys account for 4 / 7 of the total number, and the total number is regarded as 7. Boys account for 4 and girls for 3
So girls are equivalent to 3 / 4 of boys, that is, 3 / 4
Given sin а = - 2cosa, find the value of COS & # 178; a-SiN & # 178; a
A: Sina = - 2cosa is substituted into sin & # 178; a + cos & # 178; a = 1 has: 4cos & # 178; a + cos & # 178; a = 1 so: cos & # 178; a = 1 / 5 so: cos & # 178; a-SiN & # 178; a = 2cos & # 178; A-1 = 2 * (1 / 5) - 1 = - 3 / 5 so: cos & # 178; a-SiN & # 178; a = - 3 / 5
The number of male students is 4 / 5 of the number of female students. What is the number of male students?
I only want to answer 4 / 9, but I didn't expect that Du Niang would not let me write such a short answer, so I'll write a longer one, 4 / 9
Given 3sina-2cosa = 0, find the value of 3 (COSA) ^ 2-2sinacosa + 1
cosa=3/2sina
Original formula = Sina ^ 2-2sina * 3 / 2sina + 9 / 4sina ^ 2
=1/4sina^2
Cosa ^ 2 + Sina ^ 2 = 1
We get Sina ^ 2 = 4 / 13
Then the original formula is 1 / 13
The number of male students is 4 / 9 of the total number of students, and the number of female students is a fraction of the number of male students
4 out of 1-9 = 5 out of 9
5 out of 9 = 9:5
4 divided by 5
A: the number of girls is five fourths of that of boys
Then ﹣ a = ﹣ a = ﹣ a = ﹣ a = ﹣ a?
3sinA+cosA=0
tana=-1/3
cos²A+2sinAcosA=(cos²A+2sinAcosA)/(sina^2+cosa^2)=(1+2tana)/(tana^2+1)=(1-2/3)/10/9=3/10
Twelve