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faust18 [17]
3 years ago
5

If the Bills gained 4 yards on first down, then lost 6 yards on second down, then gained 10 yards on third down, how many yards

did they actually move forward on that drive? (You can pretend they started at 0 yards to make it easier​
Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
7 0

Answer:

8 yards

Step-by-step explanation:

4 - 6 = -2

-2 + 10 = 8

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How do you draw and number line and graph with -5 and -(-5) on it
NARA [144]

Answer:

see below

Step-by-step explanation:

Draw a line with points on it from -10 to 10 going by 1

Put a dot at -5 for the -5 point

-(-5) is 5

Put a dot at 5

6 0
3 years ago
Read 2 more answers
26-4=12+ missing number
saw5 [17]
22-12=10

so this missing answer after the + is 10
7 0
3 years ago
Read 2 more answers
128 POINTS IF YOU HELP
Aleksandr-060686 [28]
(A) We let y = the cost and are told x = the number of people. Since you pay $20 per person, the cost is 20x. That is, y=20x

(B) Again, let the cost =y and the number of people is given as x. You pay $10 per person or 10x plus an additional $50 for the room. That is, y=10x+50

(c) Link to graphs: https://www.desmos.com/calculator but if that doesn't work see the attachment for a screen shot. You just have to put the equations (type them) at left and the graph comes automatically.

(D) The admission price is the same when the two equations are equal. You can find this by setting them equal to each other as such: 20x = 10x+50 and solving for x. However, since you just graphed them the point of intersection (where the lines share/have the same point) gives the information. Remembers that (x,y) = (people, cost). The graphs intersect at (5, 100) so for 5 people the cost is the same and the cost is $100.

(E) For the regular rate we let x = 6 and solve for y (the cost). We get y = 20x which is y = (20)(6)=120. It costs $120 using the regular rate to take 6 people. Now let's use the equation for the group rate again with x = 6. Here we get y = 10x +50 or y = 10(6)+50 = $110. The group rate costs $110.

(F) The cost is the same at 5 people but if there are more than five the group rate is better as we saw in part E. So the regular rate is better for less than 5 people.

(G) Here y = $150. Let us use the group rate formula and solve for x (the number of people). 10x+50 = 150 so 10x = 100 and x = 10. Since 10 is more than 5 this is the better deal. However if you don't believe it or want to double check we can solve for x using y = 150 and the regular rate equation. We get: 20x = 150 so x = 7.5 Since we can't bring half a person we would only be able to bring 7 and that is less than 10 so this is not the best choice. Use the group rate and bring 10 people!

8 0
3 years ago
Read 2 more answers
Plz help me plz I need answer and explanation plz
pentagon [3]

9514 1404 393

Answer:

  A = {put, the, cake, in, basket}

  B = {is, the, cake, in, basket}

  C = {no, the, cake, in, basket}

  D = { }

Step-by-step explanation:

A = (A -B) ∪ (A∩B) ∪ (A∩C}

A = {put} ∪ {the, cake, in, basket} ∪ {the, cake, in, basket}

A = {put, the, cake, in, basket}

__

B = (B -A) ∪ (B -C) ∪ (B∩A) ∪ (B∩C)

B = {is} ∪ {is} ∪ {the, cake, in, basket} ∪ {the, cake, in, basket}

B = {is, the, cake, in, basket}

__

C = (C -B) ∪ (C∩A) ∪ (C∩B)

C = {no} ∪ {the, cake, in, basket} ∪ {the, cake, in, basket}

C = {no, the, cake, in, basket}

__

D = { } . . . . not defined anywhere

8 0
3 years ago
Explain why ΔABC cannot be shown to be congruent to ΔCDA.
igomit [66]

There isn't enough info to prove the triangles to be congruent or not. So we can't say for sure either way.

We have angle CAD = angle ACB given by the arc markings, and we know that AC = AC due to the reflexive theorem. However we are missing one third piece of information.

That third piece of info could be....

  • AD = BC which allows us to use SAS
  • angle ACD = angle CAB which allows us to use ASA
  • angle ABC = angle CDA which allows us to use AAS (slight variation of ASA)

Since we don't know any of those three facts, we simply don't have enough information.

side note: If AB = CD, then this leads to SSA which is not a valid congruence theorem. If we had two congruent sides, the angle must be between the two sides, which is what AD = BC allows.

4 0
3 years ago
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