Answer:
y = |x| - 1
Step-by-step explanation:
The difference between the parent function (y=|x|) and the graph, is that the graph is 1 unit down. y = |x| -1 has the parent function 1 unit down
Answer: x = 11
Step-by-step explanation:
10x+84=194
10x = "10 times a number"
+84 = "increased by eighty four"
=194 = "equals 194"
10x+84=194
Subtract 84
10x = 110
Divide by 10
x = 11
<em>Hope it helps <3</em>
The function will be a (linear function) linear equation in two variable and the equation of the function is y = 87x - 783
<h3>What is a linear equation?</h3>
It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.
If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.
We have given data:
Game(x): 13 14 15 16 17 18
Attendance(y): 348 435 552 609 696 783
If plot these points on a coordinate plane, we will see these points will align in a straight line.
We know we can find a line equation with two points:
(13, 348) and (14, 435)
![\rm (y - 435) = \dfrac{435-348}{14-13}(x-14)](https://tex.z-dn.net/?f=%5Crm%20%28y%20-%20435%29%20%3D%20%5Cdfrac%7B435-348%7D%7B14-13%7D%28x-14%29)
y - 435 =87(x-14)
y = 87x - 783
Thus, the function will be a (linear function) linear equation in two variable and the equation of the function is y = 87x - 783
Learn more about the linear equation here:
brainly.com/question/11897796
#SPJ1
Answer:
3200 ft-lb
Step-by-step explanation:
To answer this question, we need to find the force applied by the rope on the bucket at time ![t](https://tex.z-dn.net/?f=t)
At ![t=0, the weight of the bucket is 6+36=42 \mathrm{lb}](https://tex.z-dn.net/?f=t%3D0%2C%20the%20weight%20of%20the%20bucket%20is%206%2B36%3D42%20%5Cmathrm%7Blb%7D)
After
seconds, the weight of the bucket is ![42-0.15 t \mathrm{lb}](https://tex.z-dn.net/?f=42-0.15%20t%20%5Cmathrm%7Blb%7D)
Since the acceleration of the bucket is the force on the bucket by the rope is equal to the weight of the bucket.
If the upward direction is positive, the displacement after
seconds is ![x=1.5 t](https://tex.z-dn.net/?f=x%3D1.5%20t)
Since the well is 80 ft deep, the time to pull out the bucket is ![\frac{80}{2}=40 \mathrm{~s}](https://tex.z-dn.net/?f=%5Cfrac%7B80%7D%7B2%7D%3D40%20%5Cmathrm%7B~s%7D)
We are now ready to calculate the work done by the rope on the bucket.
Since the displacement and the force are in the same direction, we can write
![W=\int_{t=0}^{t=36} F d x](https://tex.z-dn.net/?f=W%3D%5Cint_%7Bt%3D0%7D%5E%7Bt%3D36%7D%20F%20d%20x)
Use
and ![F=42-0.15 t](https://tex.z-dn.net/?f=F%3D42-0.15%20t)
![W=\int_{0}^{36}(42-0.15 t)(1.5 d t)](https://tex.z-dn.net/?f=W%3D%5Cint_%7B0%7D%5E%7B36%7D%2842-0.15%20t%29%281.5%20d%20t%29)
![=\int_{0}^{36} 63-0.225 t d t](https://tex.z-dn.net/?f=%3D%5Cint_%7B0%7D%5E%7B36%7D%2063-0.225%20t%20d%20t)
![=63 \cdot 36-0.2 \cdot 36^{2}-0=3200 \mathrm{ft} \cdot \mathrm{lb}](https://tex.z-dn.net/?f=%3D63%20%5Ccdot%2036-0.2%20%5Ccdot%2036%5E%7B2%7D-0%3D3200%20%5Cmathrm%7Bft%7D%20%5Ccdot%20%5Cmathrm%7Blb%7D)
![=\left[63 t-0.2 t^{2}\right]_{0}^{36}](https://tex.z-dn.net/?f=%3D%5Cleft%5B63%20t-0.2%20t%5E%7B2%7D%5Cright%5D_%7B0%7D%5E%7B36%7D)
![W=3200 \mathrm{ft} \cdot \mathrm{lb}](https://tex.z-dn.net/?f=W%3D3200%20%5Cmathrm%7Bft%7D%20%5Ccdot%20%5Cmathrm%7Blb%7D)