There are 30 boys and 18 girls in a class. How many times of the class are boys and girls? I want Formula 1

There are 30 boys and 18 girls in a class. How many times of the class are boys and girls? I want Formula 1

30 + 18 = 48 30 △ 48 = five eighths
30 △ 18 ≈ 1.7 times
Hope to help you
Find Sin & sup2; 1 & ordm; + Sin & sup2; 2 & ordm; + Sin & sup2; 3 & ordm; + +sin²88º+sin²89º.
sin89°=cos1°
sin88°=cos2°
……
So the original formula = (Sin & sup2; 1 ° + cos & sup2; 1 °) + (Sin & sup2; 2 ° + cos & sup2; 2 °) +（sin²44°+cos²44°）+sin²45 °
=44.5
There are 18 boys and 22 girls. Boys account for a few percent of the class, while boys account for a few percent of girls
Boys account for 18 out of 40 girls 18 + 22 = 40 18 divided by 40 = 18 out of 40 boys are 9 out of 11 girls 18 divided by 22 = 9 out of 11
(√ (1 + 2cos260 & ordm; cos350 & ordm;) / cos10 & ordm; - (√ (1-cos & sup2; 170 °)?
Cos (5 π / 6 + x) = cos [π - (π / 6-x)] = - √ 3 / 3 √ (1 + 2cos260 degrees, cos350 degrees) / [cos10 degrees - √ (1-cos ^ 2 * 170 degrees)] = √ (1-2sin10 degrees, cos10 degrees) / [cos10 degrees - √ (1-cos ^ 2 * 170 degrees)] = (cos10 degrees - sin10 degrees) / (cos10 degrees - sin170 degrees) = (cos10 degrees - sin10 degrees) = (cos10 degrees) / (cos10 degrees - sin10 degrees) = (cos10 degrees - sin10 degrees) = (cos10 degrees) = (cos10 degrees - sin10 degrees) = (cos10 degrees)
The number of male students in Grade 6 is 7 / 10 of that of female students, so what percentage of the total number of male students in the whole year
7 out of 17 is equal to 10 girls and 7 out of 17 boys, so boys account for 7 out of 17 and 7 out of 77
seven-seventeenths
Yes
Seven out of seventeen is equal to 10 girls and 7 boys, so boys account for 7 out of 17 and 7 out of 77
17/7
Calculation (√ (1 + 2cos260 & ordm; cos350 & ordm;) / cos10 & ordm; - (√ (1-cos & sup2; 17)
(√(1+2cos260ºcos350º))/cos10º-(√(1-cos²17=(√[1+2cos(180+80º)cos(360-10º)]/cos10º-(√(1-cos²17)=(√[1-2cos80ºcos10º]/cos10º-(√(1-cos²17)=...
The number of male students in Grade 6 is 710 times that of female students, so the number of male students accounts for & nbsp; & nbsp; & nbsp; & nbsp; & nbsp;) of the whole grade
Suppose the number of boys is 7, then the number of girls is 10, the number of students in the whole grade is 7 + 10 = 17, 7 △ 17 = 717
Find the square of 2sin20 '+ 2cos5' + tan20 ` sin10 '
I also want to know the whole process of solving this problem
2sin20 '+ (square of 2cos5') + tan20'sin10 '= 2sin20' + (square of 2cos5 ') + (sin20'sin10') / cos20 '... (cosine multiple angle formula) = 2sin20' + (1 + cos10 ') + (sin20'sin10') / cos20 '... (shift) = 2sin20' + 1 + (cos20'cos10 '+ sin20'sin10'... (cosine multiple angle formula) = 2sin20 '+ (1 + cos10') / cos20 '... (shift) = 2sin20' + 1 + (cos20'cos10 '+ sin20'sin10')
The integral sum difference formula should be replaced by the angle that can get the accurate value
Then calculate
As for the formula, I can't remember it clearly
It can be calculated with a special scientific calculator for students
To change sin20 cos5, tan20 sin10 into arithmetic numbers, write them down at one time and calculate them with a calculator
The number of male students in Grade 6 is 710 times that of female students, so the number of male students accounts for & nbsp; & nbsp; & nbsp; & nbsp; & nbsp;) of the whole grade
Suppose the number of boys is 7, then the number of girls is 10, the number of students in the whole grade is 7 + 10 = 17, 7 △ 17 = 717
Can sin50 ° and cos50 ° be simplified again
√2sin85°
The original formula = √ 2 (sin50 ° * cos45 ° + cos50 ° * sin45 °)
=√2sin(50°+45°)
=√2sin95°
=√2sin85°
Of course
sin50°+cos50°
=√2(sin50°*cos45°+cos50°*sin45°)
=√2(sin(50°+45°))
=√2sin95°
=√2sin85°
=√2*sin95°
Of course