To answer this question, we let x be the the consulting fee of Iris. With this representation, the value for the hourly rate is equal to 11x. The equation that would allow us to relate the consulting services fee, hourly fee and total value given that she worked for 7 hours would then be equal to,
x + (7)(11x) = 470
Simplifying the left-hand side of the equation,
78x = 470
Dividing the equation by 78 will give us an answer of 6.
Hence,
<em> x = $6 (Consulting services fee)</em>
<em> 11x = 11($6) = $66 (hourly rate)</em>
Answer: 9
Step-by-step explanation:
To solve this problem you must apply the proccedure shown below:
Keeping on mind that:
(-)(-)=+
(+)(+)=+
(-)(+)=-
You can distribute the signs as following:
Now, you can add all the numbers shown above. Therefore, you obtain the following sum:
Answer:
t=2
Step-by-step explanation:
We are given the equation
h = -16t²+36t+1
Next, we are asked to find how long the ball was in the air (the time, in seconds) if Casey caught the ball 9 feet above the ground (9 feet is the height).
Therefore, as 9 is the height, we can plug that into our equation to get
9 = -16t²+36t+1
To make this a quadratic equation that is easy to factor, we can subtract 9 from both sides to get
-16t²+36t-8=0
To factor this, we need to find two values that add up to 36 and multiply to (-16)*(-8) = 128. With a little guess and check, I found the numbers 32 and 4 to work well. We can then make the equation
-16t²+32t+4t-8=0
-16t(t-2) + 4(t-2) = 0
(-16t+4)(t-2) = 0
Therefore, to solve for the equation being 0, or 9 = -16t²+36t+1, either (t-2) or (-16t+4) must equal 0
t-2 = 0
t=2
-16t+4 = 0
-16t = -4
t = 1/4
Therefore, t=1/4 and/or t=2. To figure out which one it is (or both!), we can take into account that the question states "on its way down". This means that the ball goes up and then down, and we want to find t when it is going down. Using the knowledge that a ball that is hit goes up and then down, as well as that -16t²+36t-8=0 is a quadratic equation with 2 solutions, we can say that the higher value of t (t=2) represents when the ball is going down
