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GREYUIT [131]
4 years ago
5

How can i write 1 as a power in three different ways?

Mathematics
1 answer:
Svetradugi [14.3K]4 years ago
5 0
2^1         1^2
3^1   or   1^3
4^1         1^4
I'm not sure which.
You might be interested in
Over what interval is the function of the graph
Svet_ta [14]

Answer:

1 ≤ x ≤ 6                             Option A

Step-by-step explanation:

When graph is constant it means, value of graph ( y-values)  not to be change different values of x in given interval.

In other words, graph must be parallel to the x axis is known as the constant graph.

A function is a measure of the rate of change of function values with respect to change in input values.

In given graph horizontal line is lie on the interval from x ≥ 1 to x ≤ 6.

So, option A is correct.

6 0
4 years ago
26. Define a relation ∼ ∼ on R 2 R2 by stating that ( a , b ) ∼ ( c , d ) (a,b)∼(c,d) if and only if a 2 + b 2 ≤ c 2 + d 2 . a2+
Tresset [83]

Answer:

~ is reflexive.

~ is asymmetric.

~ is transitive.

Step-by-step explanation:

~ is reflexive:

i.e., to prove $ \forall (a, b) \in \mathbb{R}^2 $, $ (a, b) R(a, b) $.

That is, every element in the domain is related to itself.

The given relation is $\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$

Reflexive:

$ (a, b) \sim (a, b) $ since $ a^2 + b^2 = a^2 + b^2 $

This is true for any pair of numbers in $ \mathbb{R}^2 $. So, $ \sim $ is reflexive.

Symmetry:

$ \sim $ is symmetry iff whenever $ (a, b) \sim (c, d) $ then $  (c, d) \sim (a, b) $.

Consider the following counter - example.

Let (a, b) = (2, 3) and (c, d) = (6, 3)

$ a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13 $

$ c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42 $

Hence, $ (a, b) \sim (c, d) $ since $ a^2 + b^2 \leq c^2 + d^2 $

Note that $ c^2 + d^2 \nleq a^2 + b^2 $

Hence, the given relation is not symmetric.

Transitive:

$ \sim $ is transitive iff whenever $ (a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f) $ then $ (a, b) \sim (e, f) $

To prove transitivity let us assume $ (a, b) \sim (c, d) $ and $ (c, d) \sim (e, f) $.

We have to show $ (a, b) \sim (e, f) $

Since $ (a, b) \sim (c, d) $ we have: $ a^2 + b^2 \leq c^2 + d^2 $

Since $ (c, d) \sim (e, f) $ we have: $ c^2 + d^2 \leq e^2 + f^2 $

Combining both the inequalities we get:

$ a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2 $

Therefore, we get:  $ a^2 + b^2 \leq e^2 + f^2 $

Therefore, $ \sim $ is transitive.

Hence, proved.

3 0
3 years ago
35% of x is 3.5 What is the value of x
docker41 [41]

Answer: x =10

Step-by-step explanation:

I'm sorry if its not right

4 0
3 years ago
Mitchell took a math exam today.If he got 63 out of 84 questions correct, what percentage did Mitchell get wrong?
NISA [10]

Answer:

The answer is x = 25%

Step-by-step explanation:

84 represents 100% of the questions.

63 represents x% of the correct answers.


x = (63x100)/84

x = 6300/84

x = 75% correct answers.


percentage of the wrong answers:

100% - 75% = 25%

5 0
3 years ago
Read 2 more answers
A bucket contains 50 lottery balls numbered 1-50. One is drawn at random. What is the probability that it is a multiple of 6 giv
never [62]
The probability is 7/41.

We are looking for 
P(multiple of 6 given 2 digit) = P(2 digit & multiple of 6)/P(2 digit)

There are 7 2-digit numbers that are multiples of 6 between 1 and 50, out of 50 numbers; P(2 digit & multiple of 6) = 7/50

There are 41 2-digit numbers out of 50; P(2 digit) = 41/50

This gives us 7/50 ÷ 41/50 = 7/50 × 50/41 = 7/41
3 0
3 years ago
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