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olasank [31]
2 years ago
13

(80 + 240 times 3) ÷ 20 Solve the equation

Mathematics
1 answer:
shutvik [7]2 years ago
8 0
240x3= 720

720+80 = 800

800/20= 40

The answer is 40.

Tell me if this helps
Please mark me as brainliest
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(4+7 +5+5+2) +[14 5/8+ 14 3/4 +14 7/8+15 + 15 1/8]=​
Reptile [31]

Answer:

97 and 1/8

Step-by-step explanation:

(4 + 7 + 5 + 5 + 2) + (14 5/8 + 14 3/4 + 14 7/8 + 15 + 15 1/8)

23 + (14 5/8 + 14 3/4 + 14 7/8 + 15 + 15 1/8)

23 + 74 1/8

97 and 1/8

7 0
2 years ago
Read 2 more answers
Find an equation for the nth term of the arithmetic sequence.
vekshin1
The answer is the second answer choice, an = -13 + 5(n - 1)
5 0
2 years ago
Read 2 more answers
Зу - 2(5y - 7) = -9y + 6​
Iteru [2.4K]

Answer:

y = -4

Step-by-step explanation:

3y - 2(5y - 7) = -9y + 6 (Given)

3y - 10y + 14 = -9y + 6 (Distribute)

-7y + 14 = -9y + 6 (Add like terms)

2y + 14 = 6 (Add 9y on both sides)

2y = -8 (Subtract 14 on both sides)

y = -4 (Divide 2 on both sides)

You can check by substituting the solution for y:

3(-4) - 2(5(-4) - 7) = -9(-4) + 6

-12 - 2(-20 - 7) = 36 + 6

-12 - 2(-27) = 42

-12 + 54 = 42

42 = 42

4 0
3 years ago
a. Determine all bijections from the {1,2,3} into {a,b,c}. b. Determine all bijections from {1, 2, 3} into {a,b,c,d}.
avanturin [10]

Part A

There are 6 bijections from {1,2,3} to {a,b,c}. This is effectively the same as asking the question "how many ways are there to arrange {a,b,c} where order matters?" We use a factorial to answer this question.

3 factorial = 3! = 3*2*1 = 6

You can also use a permutation, which is composed of factorials, to get the same answer.

======================================================

Part B

There are no bijections from {1,2,3} to {a,b,c,d}. Why is this? Because a bijection has two properties: it must be one-to-one, and it must be onto. The term "onto" in mathematics means "every value in the range is targeted". In the case of the range {a,b,c,d} it is not possible for each value to show up. This is because there are only three items in the domain {1,2,3}. You'll always be one letter short.

As you can probably guess, a bijection is only possible if and only if n(D) = n(R), where D and R are the domain and range respectively. The notation n(D) represents the count or number of items in set D.

3 0
3 years ago
Can someone please help I pay £100
Flura [38]

Answer:

The new points to the triangle will be:
A(1,4)\\ B(1,1)\\ C(-1,1)

Step-by-step explanation:

Because the reflection point is at x=2, all x values will subtract their distances from x=2 to get their new values. The y values remain the same.

The starting values are:

A(3,4)\\ B(3,1)\\ C(5 ,1)

Point A is 1 unit away from x=2, so we'll subtract 1 from 2 to get the new x value: 2 - 1 = 1, so A(1,4).

 

Point B is also 1 unit away from x=2, so we'll subtract 1 from 2 to get the new x value: 2 - 1 = 1, so B(1,1).

Point C is 3 units away from x=2, so we'll subtract 3 from 2 to get the new x value: 2 - 3 = -1, so C(-1,1).

3 0
2 years ago
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