7cos(x) + 1 = 6sec(x)
7cos(x) + 1 = 6/cos(x)
7cos^(x) + cos(x) = 6
7cos^(x) + cos(x) - 6 = 0
[7cos(x) - 6][cos(x) + 1] = 0
cos(x) = 6/7 , x = arccos(6/7) and
cos(x) = -1, x = 180
The answer of this question is B
Step-by-step explanation:
EF = 4x - 15
FG = 3x - 7
EG = EF + FG = 20
so,
4x - 15 + 3x - 7 = 20
7x - 22 = 20
7x = 42
x = 6
EF = 4×6 - 15 = 9
FG = 3×6 - 7 = 11
A = l x w
so you know that the length is 3m longer than the width, so you could use a formula to represent that
w = l + 3
you then substitute the second equation into the first to solve for l
70 = l x (l +3)
70 = l^2 + 3l
you could then rearrange the formula and solve for l using the quadratic formula
0 = l^2 + 3l - 70
l = -3 +- (square root (3)^2 - 4(1)(70)) / 2(1)
l = -3 +- (square root 9 + 280) / 2
l = -3 +- (square root 289) / 2
l = -3 +- 17 / 2
then you solve for the two seperate roots
l = -3 + 17 /2
l = 14 / 2
l = 7
or
l = -3 - 17 / 2
l = -20 / 2
l = -10
since a length cannot be negative, this root is not viable. therefore l = 7
to solve for w you would use
w = l + 3
w = 7 + 3
w = 10
hope this helps! if you did not understand a step or concept please let me know!
we know there are 5280 feet in 1 mile, and there are 3600 seconds in 1 hour.
