Answer:
d - 7
Step-by-step explanation:
The word "difference" tells us that we will be using subtraction. Also, subtraction is not the same if you reverse the terms like, x - 2 and 2 - x are not the same. In your question it say "d and 7" so that's the order we use in the algebraic expression.
Answer:
2 per bottle
Step-by-step explanation:
12 bottles÷ 6 dollars=2
Short answer: No and yes.
Yes First.
If the lower right point (looks like an C perhaps or an F) and the middle point and the upper left point lie on the same line and if D the middle point and the upper right point lie n the the same line, then yes. You have angle side angle. The lower right angle is given as equal to the upper left angle. The angles at C are vertically opposite angles and they are equal and you have been given 2 enclosed sides that are equal.
No
If the points I've described are not on the same line, then the triangles are not congruent. If I can, I'll edit this a second time.
Please Note:
You have done something that looks a bit illegal to me. I wouldn't just trade pictures. It only costs you another 5 points to issue another question. I do think, however, that it is very clever. In answering about 1700 questions, I've never seen this done. That's commendable.
The equation of the line that is perpendicular to 4x - 3y = 10 through the point (-2,4) is ![y-4=\frac{-3}{4}(x+2)](https://tex.z-dn.net/?f=y-4%3D%5Cfrac%7B-3%7D%7B4%7D%28x%2B2%29)
<u>Solution:</u>
Given, line equation is 4x – 3y = 10
We have to find a line that is perpendicular to 4x – 3y = 10 and passing through (-2, 4)
Now, let us find the slope of the given line,
![\text { Slope of a line }=\frac{-\mathrm{x} \text { coefficient }}{\mathrm{y} \text { coefficient }}=\frac{-4}{-3}=\frac{4}{3}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Slope%20of%20a%20line%20%7D%3D%5Cfrac%7B-%5Cmathrm%7Bx%7D%20%5Ctext%20%7B%20coefficient%20%7D%7D%7B%5Cmathrm%7By%7D%20%5Ctext%20%7B%20coefficient%20%7D%7D%3D%5Cfrac%7B-4%7D%7B-3%7D%3D%5Cfrac%7B4%7D%7B3%7D)
We know that, slope of a line
slope of perpendicular line = -1
![\begin{array}{l}{\text { Then, } \frac{4}{3} \times \text { slope of perpendicular line }=-1} \\\\ {\rightarrow \text { slope of perpendicular line }=-1 \times \frac{3}{4}=-\frac{3}{4}}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%5Ctext%20%7B%20Then%2C%20%7D%20%5Cfrac%7B4%7D%7B3%7D%20%5Ctimes%20%5Ctext%20%7B%20slope%20of%20perpendicular%20line%20%7D%3D-1%7D%20%5C%5C%5C%5C%20%7B%5Crightarrow%20%5Ctext%20%7B%20slope%20of%20perpendicular%20line%20%7D%3D-1%20%5Ctimes%20%5Cfrac%7B3%7D%7B4%7D%3D-%5Cfrac%7B3%7D%7B4%7D%7D%5Cend%7Barray%7D)
Now, slope of our required line =
and it passes through (-2, 4)
<em><u>The point slope form is given as:</u></em>
![\begin{array}{l}{y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on the line. }} \\\\ {\text { Here in our problem, } m=-\frac{3}{4}, \text { and }\left(x_{1}, y_{1}\right)=(-2,4)} \\\\ {\text { Then, line equation } \rightarrow y-4=-\frac{3}{4}(x-(-2))}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7By-y_%7B1%7D%3Dm%5Cleft%28x-x_%7B1%7D%5Cright%29%20%5Ctext%20%7B%20where%20%7D%20m%20%5Ctext%20%7B%20is%20slope%20and%20%7D%5Cleft%28x_%7B1%7D%2C%20y_%7B1%7D%5Cright%29%20%5Ctext%20%7B%20is%20point%20on%20the%20line.%20%7D%7D%20%5C%5C%5C%5C%20%7B%5Ctext%20%7B%20Here%20in%20our%20problem%2C%20%7D%20m%3D-%5Cfrac%7B3%7D%7B4%7D%2C%20%5Ctext%20%7B%20and%20%7D%5Cleft%28x_%7B1%7D%2C%20y_%7B1%7D%5Cright%29%3D%28-2%2C4%29%7D%20%5C%5C%5C%5C%20%7B%5Ctext%20%7B%20Then%2C%20line%20equation%20%7D%20%5Crightarrow%20y-4%3D-%5Cfrac%7B3%7D%7B4%7D%28x-%28-2%29%29%7D%5Cend%7Barray%7D)
![y-4=\frac{-3}{4}(x+2)](https://tex.z-dn.net/?f=y-4%3D%5Cfrac%7B-3%7D%7B4%7D%28x%2B2%29)
Hence the equation of line is found out