Answer:
2 sqrt(3) = n
k = 4 sqrt(3)
Step-by-step explanation:
We know that cos 30 = adjacent/ hypotenuse
cos 30 = 6 /k
Multiply each side by k
k cos 30 = 6k * k
k cos 30 = 6
Divide each side by cos 30
k cos 30/ cos 30 = 6/ cos 30
k = 6 / cos 30
k = 6/ (sqrt(3)/2)
k =12 / sqrt(3)
We cannot leave the sqrt in the denominator
k = 12 / sqrt(3) * sqrt(3)/sqrt(3)
k = 12 sqrt(3)/3
k = 4 sqrt(3)
We know tan 30 = opposite/adjacent
tan 30 = n/6
Multiply each side by 6
6 tan 30 = n
6 * (sqrt(3)/3) = n
6/3 * sqrt(3) = n
2 sqrt(3) = n
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Given:
Seven times the smaller of two consecutive even integers is the same as -300 minus 4 times the larger integer.
To find:
The integers
.
Solution:
Let the smaller even integer be x.
So, larger even integer is x+2 because they are consecutive even integers.
Seven times the smaller integer = 7x
-300 minus 4 times the larger integer = -300-4(x+2)
Now,




Divide both sides by 11.


So, the smaller even integer is -28.

So, the larger even integer is -26.