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Delicious77 [7]
4 years ago
13

heeeelllllpppppp heyyy i need help asap and please try your best and solve the problem out it would be greatly appreciated becau

se i have asked 5 questions already and there all wrong and no explanation at all and it needs to have ^for example, this is an exmample example 10^-1

Mathematics
2 answers:
andreev551 [17]4 years ago
7 0

Answer:

{10}^{ - 14}

Step-by-step explanation:

\huge\frac{ {10}^{ - 9} }{ {10}^{5} }  \\  \\  \huge =  {10}^{ - 9}  \times  {10}^{ - 5}  \\  \\  \huge =  {10}^{ - 9  + ( - 5)}  \\  \\ \huge =  {10}^{ - 9 - 5}  \\  \\  \huge \orange {=  {10}^{ - 14}}

Vinvika [58]4 years ago
5 0
10^-9/10^5

Simplify the expression.
Or

Reduce the expression, if possible, by cancelling the common factors.

Same answer given for both ^^

Exact form:

1/100000000000000 (14 -zeros)

Decimal form:

1 • 10^-14


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Step-by-step explanation:

the tip would be added to the meal cost

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Barack walks 2 kilometers south and then x kilometers east. ​ ​He ends 5 kilometers away from his starting position. ​ ​ Find x
amid [387]

Answer:

4.6 km

Step-by-step explanation:

-This is a Pythagorean Theorem problem where the distances walked are effectively the base and height of the imaginary right triangle thereof.

#Apply Pythagorean theorem to solve:

b^2+h^2=H^2\\\\2^2+x^2=5^2\\\\x^2=5^2-2^2=21\\\\x=\sqrt{21}\\\\x=4.58\  km

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5 0
3 years ago
3. (6 points) Determine whether the relation R on the set A is an equivalence relation a. (3 pts) A = {1,2,3,4, 5) R={(1,1), (1,
dezoksy [38]

Answer:

a is not an equivalence relation.

b is an equivalence relation.

Step-by-step explanation:

a.

A = {1,2,3,4, 5) R={(1,1), (1,2), (1,3), (2,2), (2,3), (3,1), (3,2), (3,3), (4,4), (5,5)

To see if is an equivalence relation you need to see if you have these 3 things:

Part 1: xRx for all x in A. This is the reflexive property.

Do we? Yes we have all these points in R: (1,1), (2,2) ,(3,3) ,(4,4), and (5,5).

Part 2: If xRy then yRx. This is the symmetic property.

Do we? We have (1,2) but not (2,1). So it isn't symmetric.

Part 3: If xRy and yRz then xRz.

Do we? We are not going to check this because there is no point. We have to have all 3 parts fot it be an equivalence relation.

b.

A = {a, b, c} R={(a, a), (a, c), (b, b), (c, a), (c, c)}

To see if is an equivalence relation you need to see if you have these 3 things:

Part 1: xRx for all x in A. This is the reflexive property.

Do we? Yes we have all these points in R: (a,a),(b,b), and (c,c).

Part 2: If xRy then yRx. This is the symmetric property.

Do we? We have (a,c) and (c,a). We don't need to worry about any other (x,y) since there are no more with x and y being different. This is symmetric.

Part 3: If xRy and yRz then xRz.

Do we? We do have (a,c), (c,a), and (a,a).

We do have (c,a), (a,c), and (c,c).

So it is transitive.

Question b has all 3 parts so it is an equivalence relation.

4 0
3 years ago
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