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Mandarinka [93]
3 years ago
11

Mike jogged 6 laps around a 0.25 mile track on Monday and 7 laps on Tuesday. How many miles did he jog on Monday and Tuesday com

bined? miles
Mathematics
2 answers:
frez [133]3 years ago
4 0
7 x 0.25 = 1.75
6 x 0.25 = 1.5
1.75 + 1.5 = 3.25
I hope this helps you :)
crimeas [40]3 years ago
3 0

Answer: 3.25 miles

Step-by-step explanation:

6 + 7 = 13

13 x .25 = 3.25

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Simply the problem.
Marat540 [252]

Answer:

C

Step-by-step explanation:

\sqrt{3}\cdot\sqrt{21}= \\\\\sqrt{3\cdot 21}= \\\\\sqrt{63}= \\\\3\sqrt{7}

Therefore, the correct answer is choice C. Hope this helps!

6 0
3 years ago
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Please help with this simple question!!!
strojnjashka [21]
456- 12= 444
37×12= 444
=12
7 0
2 years ago
Find the value of the power. Please do all four questions for 15 pts #1 6³ #2 9² #3 3⁴ #4 18²
defon

Answer:

6³ = 216

9² = 81

3⁴ = 81

18² = 324

Step-by-step explanation:

6³ = 6 · 6 · 6 = 36 · 6 = 216

9² = 9 · 9 = 81

3⁴ = 3 · 3 · 3 · 3 = 9 · 3 · 3 = 27 · 3 = 81

18² = 18 · 18 = 324

3 0
3 years ago
X + 1/2 = 17 need help fast
Paha777 [63]

Answer:

x=16.5

Step-by-step explanation:

-1/2 on both sides

17-0.5 or 1/2=16.5

7 0
2 years ago
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How many equivalence relations are there on the set 1, 2, 3]?
Alex787 [66]

Answer:

We need to find how many number of equivalence relations are on the set {1,2,3}

A relation is an equivalence relation if it is reflexive, transitive and symmetric.

equivalence relation R on {1,2,3}

1.For reflexive, it must contain (1,1),(2,2),(3,3)

2.For transitive, it must satisfy: if (x,y)∈R then (y,x)∈R

3. For symmetric, it must satisfy: if (x,y)∈R,(y,z)∈R then (x,z)∈R

Since (1,1),(2,2),(3,3) must be there is R, (1,2),(2,1),(2,3),(3,2),(1,3),(3,1). By symmetry,

we just need to count the number of ways in which we can use the pairs (1,2),(2,3),(1,3) to construct equivalence relations.

This is because if (1,2) is in the relation then (2,1) must be there in the relation.

the relation will be an equivalence relation if we use none of these pairs (1,2),(2,3),(1,3) . There is only one such relation: {(1,1),(2,2),(3,3)}

we can have three possible equivalence relations:

{(1,1),(2,2),(3,3),(1,2),(2,1)}

{(1,1),(2,2),(3,3),(1,3),(3,1)}

{(1,1),(2,2),(3,3),(2,3),(3,2)}

6 0
3 years ago
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