Step-by-step explanation:
-22/5 +4/5
The lcm of the denominators is 5
If you divide through u have - 22+4=-18
Answer:

Step-by-step explanation:
Write a function whose k(x) values are 5 more than four times the principal square root of x.
So,
the principal square root of x 
four times the principal square root of x 
5 more than four times the principal square root of x 
Thus,

Answer:
About 53.13 degrees.
Step-by-step explanation:
If we draw the path of the hiker, or rather, all three vectors with vector B having its starting point at the end point of vector A, a right triangle is formed. We will most likely be using trigonometric functions to solve this.
vector A is parallel to the x axis (West - East), and vector B is perpendicular, so we'll be using the inverse tangent function to get the angle based on the ratio of the lengths of vector B to vector A, which is 
atan
degrees.
If one would be pedantic, they'd say that the direction we have to go to get from vector C to vector A is clockwise, which is anti-mathematical, the true answer would be -53.13 degrees. Alternatively, whatever 360-53.13 is.
Answer:
6x+19
Step-by-step explanation:
Distribute the equation and then solve. Add/subtract the like terms
The total if all 236 coins were nickels would be $11.80, which is $3.95 short of the actual amount.
Replacing a nickel with a dime adds $0.05 to the total value, so there must have been $3.95/$0.05 = 79 such replacements.
There are 79 dimes.
There are 236 -79 = 157 nickels.
_____
Using the given variables, the problem statement gives rise to two equations. One is the based on the number of coins. The other is based on their value.
- n + d = 236
- .05n +.10d = 15.75
Solving the first for n, we get
... n = 236 - d
Substituting that into the second equation, we have
... .05(236 -d) +.10d = 15.75
... .05d = 15.75 -236·.05 . . . . . collect terms, subtract .05·236
... d = 3.95/.05 . . . . . . . . . . . . . divide by .05
... d = 79
... n = 236-79 = 157
___
The solution should look familiar, as it matches the verbal description at the beginning.