In my opinion, Jill is is correct in disagreeing to Jim. Axes on a graph do not necessarily have to begin with 0. It depends on how large are the values of what you are going to plot. If the values are too large then it is not practical to start with zero.<span />
45% right if .5 x 90 = 45 and as a percentage it would be 45%
The first step for solving this expression is to write the number below ㏒ in exponential form.
㏒
![_{ 3^{2} }](https://tex.z-dn.net/?f=%20_%7B%203%5E%7B2%7D%20%7D%20)
(x) + 2㏒
![_{ 3^{2} }](https://tex.z-dn.net/?f=%20_%7B%203%5E%7B2%7D%20%7D%20)
(y) - ㏒
![_{2}](https://tex.z-dn.net/?f=%20_%7B2%7D%20)
(z)
Using ㏒
![_{a^{y} }](https://tex.z-dn.net/?f=%20_%7Ba%5E%7By%7D%20%7D%20)
(b) =
![\frac{1}{y}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7By%7D%20)
× ㏒
![_{a}](https://tex.z-dn.net/?f=%20_%7Ba%7D%20)
(b),, transform the expressions with ㏒
![_{ 3^{2} }](https://tex.z-dn.net/?f=%20_%7B%203%5E%7B2%7D%20%7D%20)
.
![\frac{1}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20)
× ㏒
![_{3}](https://tex.z-dn.net/?f=%20_%7B3%7D%20)
(x) +
![\frac{1}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20)
× ㏒
![_{3}](https://tex.z-dn.net/?f=%20_%7B3%7D%20)
(y) - ㏒
![_{2}](https://tex.z-dn.net/?f=%20_%7B2%7D%20)
(z)
Lastly,, reduce the numbers with 2 to find your final answer.
![\frac{1}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20)
× ㏒
![_{3}](https://tex.z-dn.net/?f=%20_%7B3%7D%20)
(x) + ㏒
![_{3}](https://tex.z-dn.net/?f=%20_%7B3%7D%20)
(y) - ㏒
![_{2}](https://tex.z-dn.net/?f=%20_%7B2%7D%20)
(z)
This means that the correct answer to your question is
![\frac{1}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20)
× ㏒
![_{3}](https://tex.z-dn.net/?f=%20_%7B3%7D%20)
(x) + ㏒
![_{3}](https://tex.z-dn.net/?f=%20_%7B3%7D%20)
(y) - ㏒
![_{2}](https://tex.z-dn.net/?f=%20_%7B2%7D%20)
(z).
Let me know if you have any further questions.
:)
Answer:
Area = 4.5 in.²
Step-by-step explanation:
Area of ∆BAC = 8 in.²
EF = 3 in.
BC = 4 in.
Since both triangles are similar, therefore,
Area of ∆BAC : Area of ∆EDF = BC² : EF²
Plug in the values into the equation
8/area of ∆EDF = 4²/3²
Area of ∆EDF = (8*3²)/4²
= 4.5 in²
Answer:
depends on what the coupon is.
Step-by-step explanation: