C
pls dont get mad if its wrong
Answer:
Let two consecutive multiples of 3 be x and (x+3)
A/q,
x * (x+3) = 648
➡ x² + 3x = 648
➡ x² + 3x -648 = 0
➡ x² + 27x - 24x -648 = 0
➡ x ( x + 27 ) -24 ( x +27)
➡ ( x - 24) ( x + 27)
➡ x = 24 and x = -27
so, we take x = 24.
Required multiples of 3
➡ x = 24
➡ x +3 = 24+3 = 27.
Answer:
B
Step-by-step explanation:
b is correct
5. 6√3
6. x = 10√3 y = 30
7. x = 34 y = 17√3
8. 30
9. SinA = 3/5 CosA = 4/5
10. Tan20 = 9/x
Multiply both sides by x to get it on the other side
x(tan20) = 9
Divide 9 by tan20 to get x.
x = 9/tan20
x = 24.7
