Jason worked 20 hours at $5 per hour and 30 hours at $6 per hour.
20 hours at $5 per hour = 20 times 5

30 hours at $6 per hour = 30 times 6


He earned $280 in 50 hours. Divide:

Your answer is the last choice, $5.60
Answer:
39
Step-by-step explanation:
(16-2w)^2 + (3y÷3z)
Let w=5 y=9 and z=3
(16-2*5)^2 + (3*9÷3*3)
PEMDAS says parentheses first
Multiply and divide in the parentheses
(16-10)^2 + (27÷9)
Then add and subtract in the parentheses
(6)^2 + (3)
Now the exponent
36 +3
39
Answer:
The slope is 2.
Step-by-step explanation:
In order to find the slope of the line, we need to start by using the points in the slope formula.
m(slope) = (y2 - y1)/(x2 - x1)
m = (20 - 10)/(6 - 1)
m = 10/5
m = 2
So the slope of this line would be 2
Answer:
a) 
b) See Below for proper explanation
Step-by-step explanation:
a) The objective here is to Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the power series expansion of the given function.
The function is 
The expansion is of
is 
The expansion of cos x is 
Therefore; ![e^x + 3 \ cos \ x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ... 3[1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...]](https://tex.z-dn.net/?f=e%5Ex%20%2B%203%20%5C%20cos%20%5C%20x%20%20%3D%201%20%2B%20%5Cdfrac%7Bx%7D%7B1%21%7D%2B%20%5Cdfrac%7Bx%5E2%7D%7B2%21%7D%2B%20%5Cdfrac%7Bx%5E3%7D%7B3%21%7D%20%2B%20...%203%5B1%20-%20%5Cdfrac%7Bx%5E2%7D%7B2%21%7D%2B%20%5Cdfrac%7Bx%5E4%7D%7B4%21%7D-%20%5Cdfrac%7Bx%5E6%7D%7B6%21%7D%2B%20...%5D)

Thus, the first three terms of the above series are:

b)
The series for
is 
let consider the series; 

Thus it converges for all value of x
Let also consider the series 
It also converges for all values of x
Answer: It does not matter whether you multiply the radicands or simplify each radical first. You multiply radical expressions that contain variables in the same manner. As long as the roots of the radical expressions are the same, you can use the Product Raised to a Power Rule to multiply and simplify.
Step-by-step explanation: