- 4(1/2)x - (3/7) = (1/4)
- 4(1/2)x = (1/4) + (3/7)
- (9/2)x = (7+12)/28
-(9/2)x = 19/28
x = (19/28) * 2 / 9
x = (19/14) / 9
x = 19 / (14*9)
x = 19 / 126
Answer:

Step-by-step explanation:
We want to solve for x in 
You need to group and combine like terms and write in standard form:


By comparing to
, we a=1,b=-2 and c=-46
The solution can be obtained using the quadratic formula.

We substitute the coefficients to get:





The last choice is correct
Given age, one can not determine the height. In another words, students with same age may have different height
Step-by-step explanation:
Left hand side:
tan 203° + tan 22° + tan 203° · tan 22°
From angle sum identities, we know:
tan (α + β) = (tan α + tan β) / (1 − tan α · tan β)
Therefore:
tan (203° + 22°) = (tan 203° + tan 22°) / (1 − tan 203° · tan 22°)
tan (225°) = (tan 203° + tan 22°) / (1 − tan 203° · tan 22°)
1 = (tan 203° + tan 22°) / (1 − tan 203° · tan 22°)
tan 203° + tan 22° = 1 − tan 203° · tan 22°
Substituting:
1 − tan 203° · tan 22° + tan 203° · tan 22°
1
Answer:
(A) 165
(B) 330
Step-by-step explanation:
Total number of students in the class = 11
(A) How many different combinations of 3 selected students can he create?
ⁿCₓ = n! ÷ [(n - x)! x!]
n = 11, x = 3
11! / [8! 3!] = 165
<em>HINT: 8! or 8 factorial represents [8x7x6x5x4x3x2x1]</em>
(B) How many different combinations of 4 selected students can he create?
ⁿCₓ = n! ÷ [(n - x)! x!]
n = 11, x = 4
11! / [7! 4!] = 330
<em>Same hint applies here, for all numbers with the factorial sign.</em>