Given that Joseph has 24 cousins, and he has 6 more females than males, the number of females will be obtained as follows;
suppose he had x males, number of females will be x+6.
The total will be:
x+(x+6)=24
x+x+6=24
2x+6=24
2x=24-6
2x=18
x=18/2
x=9
thus the number of male cousins is 9, the number of female cousins will be 9+6=15
We therefore conclude that he has 9 male cousins and 15 female cousins
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<span>f(x) = 6 – 5x and g(x) = 4x – 1
f(x)-g(x) = 6-5x-(4x-1) = 6-5x-4x+1 = -9x+7
whene x = - 2
f(x)-g(x) = -9(-2)+7 = 25......(answer : D)</span>
Answer: option d. x = 3π/2Solution:function y = sec(x)
1) y = 1 / cos(x)
2) When cos(x) = 0, 1 / cos(x) is not defined
3) cos(x) = 0 when x = π/2, 3π/2, 5π/2, 7π/2, ...
4) limit of sec(x) = lim of 1 / cos(x).
When x approaches π/2, 3π/2, 5π/2, 7π/2, ... the limit →+/- ∞.
So, x = π/2, x = 3π/2, x = 5π/2, ... are vertical asymptotes of sec(x).
Answer: 3π/2
The figures attached will help you to understand the graph and the existence of multiple asymptotes for y = sec(x).
The painter painted 11 apartments because 4 doesn’t go into 45, so the closest think to 45 is 4 x 11 (44). the painter will have 1 gallon left.