I do believe it is -65d because the equation will look like this in the end
-65d=-65d.
Answer:
Jeff's Popcorn 130^3
Step-by-step explanation
(Jeff's 130^3) .
First 20.5*10^2 which is 20.5*100 which equals 2050.
Then find the volume for the other box of popcorn.
30*8^2 which is 1920.
subtract 2050 by 1920 and you get 130.
Answer:
A). Scale factor = 
B). Length of the unknown side = 2.75 feet
Step-by-step explanation:
Part A
Since, the large rectangle was scaled by a scale factor 'k' to form the smaller rectangle,
Scale factor = 
= 
= 
= 
Part B
Scale factor = 

Therefore, side length of the unknown side =
= 2.75 feet
Answer: x ≥ 7
Step-by-step explanation:
You have to square both sides of the equation. You can manipulate an equation or function any way imaginable as long as it is done equally and helps the problem become easier. If we square this, the radical will cancel and we will get two trinomials equal to each other.

We can see the two expressions are exactly equal to each other. All real numbers are solutions, besides a few. These few are when the radical is equal to a negative number. So lets set up an inequality stating that the radical must be greater than or equal to 0.

To find the zeros of a quadratic fiunction given the equation you can use the next quadratic formula after equal the function to 0:
![\begin{gathered} ax^2+bx+c=0 \\ \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20ax%5E2%2Bbx%2Bc%3D0%20%5C%5C%20%20%5C%5C%20x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5Cend%7Bgathered%7D)
For the given function:

![x=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(2)(-3)}}{2(2)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%28-10%29%5Cpm%5Csqrt%5B%5D%7B%28-10%29%5E2-4%282%29%28-3%29%7D%7D%7B2%282%29%7D)
![x=\frac{10\pm\sqrt[]{100+24}}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B10%5Cpm%5Csqrt%5B%5D%7B100%2B24%7D%7D%7B4%7D)
![\begin{gathered} x=\frac{10\pm\sqrt[]{124}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2\cdot2\cdot31}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2^2\cdot31}}{4} \\ \\ x=\frac{10\pm2\sqrt[]{31}}{4} \\ \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D%5Cfrac%7B10%5Cpm%5Csqrt%5B%5D%7B124%7D%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20x%3D%5Cfrac%7B10%5Cpm%5Csqrt%5B%5D%7B2%5Ccdot2%5Ccdot31%7D%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20x%3D%5Cfrac%7B10%5Cpm%5Csqrt%5B%5D%7B2%5E2%5Ccdot31%7D%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20x%3D%5Cfrac%7B10%5Cpm2%5Csqrt%5B%5D%7B31%7D%7D%7B4%7D%20%5C%5C%20%20%5Cend%7Bgathered%7D)
![\begin{gathered} x_1=\frac{10}{4}+\frac{2\sqrt[]{31}}{4} \\ \\ x_1=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x_1%3D%5Cfrac%7B10%7D%7B4%7D%2B%5Cfrac%7B2%5Csqrt%5B%5D%7B31%7D%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20x_1%3D%5Cfrac%7B5%7D%7B2%7D%2B%5Cfrac%7B%5Csqrt%5B%5D%7B31%7D%7D%7B2%7D%20%5Cend%7Bgathered%7D)
![\begin{gathered} x_2=\frac{10}{4}-\frac{2\sqrt[]{31}}{4} \\ \\ x_2=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x_2%3D%5Cfrac%7B10%7D%7B4%7D-%5Cfrac%7B2%5Csqrt%5B%5D%7B31%7D%7D%7B4%7D%20%5C%5C%20%20%5C%5C%20x_2%3D%5Cfrac%7B5%7D%7B2%7D-%5Cfrac%7B%5Csqrt%5B%5D%7B31%7D%7D%7B2%7D%20%5Cend%7Bgathered%7D)
Then, the zeros of the given quadratic function are:
![\begin{gathered} x=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \\ \\ x_{}=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D%5Cfrac%7B5%7D%7B2%7D%2B%5Cfrac%7B%5Csqrt%5B%5D%7B31%7D%7D%7B2%7D%20%5C%5C%20%20%5C%5C%20x_%7B%7D%3D%5Cfrac%7B5%7D%7B2%7D-%5Cfrac%7B%5Csqrt%5B%5D%7B31%7D%7D%7B2%7D%20%5Cend%7Bgathered%7D)
Answer: Third option