Figure how much percent she reads daily. Then add on till 7 until ur done... For example 7= 35% 14(days)= 70
By analyzing and understanding the graph of the absolute value function, we find that the function evaluated at the x-value equal to 1 is equal to the y-value equal to 3.
<h3>What is the y-value associated to a given x-value of an absolute value function? </h3>
In this problem we find the representation of an absolute value function, where the horizontal axis corresponds to the values of the domain, whereas the vertical axis is for the values of the range. In that picture we must look up for the y-value associated with a given x-value.
Then, we proceed to evaluate the absolute value function at x = 1. In accordance with the graph, the y-value , that is, from the vertical axis, associated with the x-value, that is, from the horizontal axis, equal to 1 is equal to a value of 3.
To learn more on absolute values: brainly.com/question/1301718
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let's first off notice that, on the 2), the sector is really half of the whole circle, and on 3) the sector is one quarter of the whole circle.
now, on 2) AB is the diameter of 4 units, therefore it has a radius of 2, or half that.
![\bf \boxed{2} \\\\\\ \stackrel{\textit{area of a circle}}{A=\pi r^2}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=2 \end{cases}\implies A=\pi 2^2\implies A=4\pi \\\\\\ \stackrel{\textit{half of that}}{A=2\pi}\implies A=\stackrel{\textit{rounded up}}{A=6.3~ft^2} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cboxed%7B2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20circle%7D%7D%7BA%3D%5Cpi%20r%5E2%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D2%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%202%5E2%5Cimplies%20A%3D4%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bhalf%20of%20that%7D%7D%7BA%3D2%5Cpi%7D%5Cimplies%20A%3D%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7BA%3D6.3~ft%5E2%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \boxed{3} \\\\\\ \stackrel{\textit{area of a circle}}{A=\pi r^2}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=20 \end{cases}\implies A=\pi 20^2\implies A=400\pi \\\\\\ \stackrel{\textit{one quarter of that}}{A=100\pi }\implies \stackrel{\textit{rounded up}}{A=314.2~in^2}](https://tex.z-dn.net/?f=%5Cbf%20%5Cboxed%7B3%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20circle%7D%7D%7BA%3D%5Cpi%20r%5E2%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D20%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%2020%5E2%5Cimplies%20A%3D400%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bone%20quarter%20of%20that%7D%7D%7BA%3D100%5Cpi%20%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7BA%3D314.2~in%5E2%7D)
Please find attached photograph. The answers are 9a^2 and 15ab.
Answer: y = -5x + 20
Step-by-step explanation:
Remember that : y=mx+b
y = (-5)x + b
Now you can plug your coordinates (-7,8) into that equation.
8 = (-5)-7 + b
Solve for b
8 = -12 + b
Add +12 to both sides of your equation:
20 = b
Now, your new equation is:
y = -5x + 20
