Answer:
x=40
Step-by-step explanation:
4x-20+x=180
5x-20=180
5x=200
x=200/5
x=40
To ease your problem, consider "L" as you x-axis
Then the coordinate become:
A(- 4 , 3) and B(1 , 2) [you notice that just the y's changed]
This is a reflection problem.
Reflect point B across the river line "L" to get B', symmetric of B about L.
The coordinates of B'(1 , -1) [remember L is our new x-axis]
JOIN A to B' . AB' intersect L, say in H
We have to find the shortest way such that AH + HB = shortest.
But HB = HB' (symmetry about L) , then I can write instead of
AH + HB →→ AH + HB'. This is the shortest since the shortest distance between 2 points is the straight line and H is the point requiered
Please mark me branliest!
**Answer**
16 if you want exact, 20 if you want a little over.
**Explanation**
So check this... it's $50 per student that attends the camp, right?
(Nod if you're with me.)
The dude buys 5 dozen donuts for 16 days. 50 x 16 = 800
5 dozen is the same as 50 donuts.
You multiply to get the total amount of donuts he buys in the total days (16).
Each 10 donuts costs 10 dollars.
If the dude buys 800 donuts it cost him $800.
And it costs 50 dollars for someone to go to camp.
So you divide 800/50 = 16 students
Now the problem says "In order to break even" I'm not sure what that means. But I'm assuming it doesn't want 16 then, because the price of 16 students would be the same (even) as the price of donuts that Mr. Brook buys. In that case it would be 20.
You are very welcome!!
Answer: a. p = c/5 - 43 b. 17 people
Step-by-step explanation:
c= 5p + 215
A) a said solve for p so we will solve for p in the equation.
c= 5p + 215 First Subtract 215 from both sides
-215 -215
c - 215 = 5p Now divide both sides by 5.
p = c/5 - 43
B) If c is the total cost of hosting a birthday party then we will input 300 into the equation for c and solve for p.
300 = 5p + 215 First subtract 215 from both sides
-215 -215
85 = 5p Divide both sides by 5
p = 17
This means 17 people can attend the meeting if Allies parents are willing to spend $300.
Look at it on the Internet