Both 15 and 20 can be divided evenly by 5, so 5 is the greatest common monomial: 5(3x^2 + 4x) or 5x(3x + 4)
Answer:
See attached
Step-by-step explanation:
[] I gave you the explanation of going from table to graph on the last question, so I will explain graph to table.
[] First, we will plot (0, 5) as it is the y-intercept.
-> I am using a square = 2-units scale as 5 is a larger number and we can see the table goes by 2's.
-> Then, using rise-over-run for the slope, we will move up one point and 2 points to the right. Repeat. Then we will graph the line.
[] Now, we can put (0, 5) into the table.
-> For the rest, we will look at our graph.
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.
- Heather
Answer:
Solution given:
BD=35ft
AC=43ft
AB=8ft
EF=33-8=25ft
AE=26ft
Now
area of right angled triangle ∆AEF
=½(EF*AE)=½(25×25)=312.5ft²
area of parallel trapezoid=½AB(BD+AC)
=½*8(35+43)=312ft²
<u>So</u>
<u>Area</u><u> </u><u>of</u><u> </u><u>polygon</u><u> </u><u>:</u><u> </u><u>3</u><u>1</u><u>2</u><u>.</u><u>5</u><u>+</u><u>3</u><u>1</u><u>2</u><u>=</u><u>6</u><u>2</u><u>4</u><u>.</u><u>5</u><u>f</u><u>t</u><u>²</u><u> </u><u>is</u><u> </u><u>a</u><u> </u><u>required</u><u> </u><u>answer</u><u>.</u>
Answer:
See below ~
Step-by-step explanation:
<u>Equation Plan A</u>
<u>Equation Plan B</u>
<u>Equating them</u>
- 40x + 860 = 100x + 200
- 60x = 660
- x = 11
- The plans will pay the same at 11 correct questions.
<u>Solving for y using x = 11</u>
- y = 100(11) + 200
- y = 1100 + 200
- y = 2300
- The amount they'll pay at that number of questions is $2,300
<u>Comparing both plans when x = 13</u>
- Plan A : 40(13) + 860 = 520 + 860 = $1,380
- Plan B : 100(13) + 200 = 1300 + 200 = $1,500
- As Plan B pays more, that is the plan you should choose
Answer: Choice 4) A to B to C to E to D to A to E to B
An Euler path is a path where you use each edge or road exactly one time. Think of it like crossing a bridge and when you cross the bridge, the rope is cut so you can't reuse the bridge again. If we started at A, went to B, then to C, then to E, then to D, then back to A, then to E again, then finally to B, we would use up all of the edges. I've provided a diagram of what's going on (see attached image below). The idea is to start at point A, follow the red arrows until you get back to point A, then follow the blue arrows until you get to point B. This is <u>not</u> a circuit because we started at point A and ended up at some other point that isn't point A (in this case, point B). We would need to end up at the same starting location to have a circuit.