Welming's spending is $1164.8 and his savings is $291.2
<h3>How to determine the savings and the spending?</h3>
The given parameters are:
Weekly pocket = $28
Save = 20%
There are 52 weeks in a year.
So, the yearly pocket is:
Yearly pocket = $28 * 52
Evaluate
Yearly pocket = $1456
He saves 20%.
So, we have:
Savings = 20% * $1456
Evaluate
Savings = $291.2
His spending is then calculated as:
Spending = $1456 - $291.2
Evaluate
Spending = $1164.8
Hence, Welming's spending is $1164.8 and his savings is $291.2
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Answer:
I think that it is y=15x+35
Step-by-step explanation:
15*x is how many hours and you have to pay $35 regardless to rent the boat
There are two ways to list the angles:
1) Simply name them based on the points:
∠W , ∠X , ∠Y , ∠Z
2) The way that I believe that you are supposed to list them in this case. List them as such:
∠WYZ , ∠YZX , ∠ZXW , ∠XWY
~
1. Know what you're looking for:
The range is the difference between the lowest and highest values.
Your weight is 145.2 and can vary from the actual weight by maximum 0.3 lbs less or maximum 0.3 lbs more.
2. Calculate your lowest and highest values :
Lowest : 145.2 - 0.3 = 144.9 lbs
Highest : 145.2 + 0.3 = 145.5 lbs
3. Calculate the range of actual weights of the object :
145.5 - 144.9 = 0.6 lbs
--> The answer is: <u>The range of actual weights of the object is 0.6 lbs.</u>
There you go! I really hope this helped, if there's anything just let me know! :)
A.) To find the maximum height, we can take the derivative of h(t). This will give us the rate at which the horse jumps (velocity) at time t.
h'(t) = -32t + 16
When the horse reaches its maximum height, its position on h(t) will be at the top of the parabola. The slope at this point will be zero because the line tangent to the peak of a parabola is a horizontal line. By setting h'(t) equal to 0, we can find the critical numbers which will be the maximum and minimum t values.
-32t + 16 = 0
-32t = -16
t = 0.5 seconds
b.) To find out if the horse can clear a fence that is 3.5 feet tall, we can plug 0.5 in for t in h(t) and solve for the maximum height.
h(0.5) = -16(0.5)^2 + 16(-0.5) = 4 feet
If 4 is the maximum height the horse can jump, then yes, it can clear a 3.5 foot tall fence.
c.) We know that the horse is in the air whenever h(t) is greater than 0.
-16t^2 + 16t = 0
-16t(t-1)=0
t = 0 and 1
So if the horse is on the ground at t = 0 and t = 1, then we know it was in the air for 1 second.
