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Simora [160]
3 years ago
7

Solve the exponential equation for x 64^7x-8 = 1

Mathematics
2 answers:
SOVA2 [1]3 years ago
7 0

X=9/64^7

Hope this helps

sashaice [31]3 years ago
6 0

Answer:

2.4 x 10^-12

Step-by-step explanation:

that is the explanation for the question

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Can you solve this problem please im having trouble with this question ​
Degger [83]

Answer:

The answer is 3.7

Step-by-step explanation:

Use cross multiplication to set up the equation then simplify,

17y = 63

  y = 63/17

  y = 3.705

Round to the nearest tenth.

Solution is y = 3.07

Good luck!

3 0
3 years ago
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23 points to whoever can answer this question !!
yKpoI14uk [10]

I do believe that that point is -2.2 because of the fact that the difference between -1.6 and -2.4 is 0.8, and there are 4 marks in between them. So each mark's value is decreased by -0.2.

5 0
3 years ago
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A perfect square ends with the same two digits. How many possible values of this digit are there?
Alex73 [517]

Answer:

A perfect square is a whole number that is the square of another whole number.

n*n = N

where n and N are whole numbers.

Now, "a perfect square ends with the same two digits".

This can be really trivial.

For example, if we take the number 10, and we square it, we will have:

10*10 = 100

The last two digits of 100 are zeros, so it ends with the same two digits.

Now, if now we take:

100*100 = 10,000

10,000 is also a perfect square, and the two last digits are zeros again.

So we can see a pattern here, we can go forever with this:

1,000^2 = 1,000,000

10,000^2 = 100,000,000

etc...

So we can find infinite perfect squares that end with the same two digits.

7 0
3 years ago
(a) Let R = {(a,b): a² + 3b <= 12, a, b € z+} be a relation defined on z+)
grin007 [14]

Answer:

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Step-by-step explanation:

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

The order to options required to show that R is an equivalence relation are;

((a, b), (a, b)) ∈ R since a·b = b·a

Therefore, R is reflexive

If ((a, b), (c, d)) ∈ R then a·d = b·c, which gives c·b = d·a, then ((c, d), (a, b)) ∈ R

Therefore, R is symmetric

If ((c, d), (e, f)) ∈ R, and ((a, b), (c, d)) ∈ R therefore, c·f = d·e, and a·d = b·c

Multiplying gives, a·f·c·d = b·e·c·d, which gives, a·f = b·e, then ((a, b), (e, f)) ∈R

Therefore R is transitive

From the above proofs, the relation R is reflexive, symmetric, and transitive, therefore, R is an equivalent relation.

Reasons:

Prove that the relation R is reflexive

Reflexive property is a property is the property that a number has a value that it posses (it is equal to itself)

The given relation is ((a, b), (c, d)) ∈ R if and only if a·d = b·c

By multiplication property of equality; a·b = b·a

Therefore;

((a, b), (a, b)) ∈ R

The relation, R, is reflexive.

Prove that the relation, R, is symmetric

Given that if ((a, b), (c, d)) ∈ R then we have, a·d = b·c

Therefore, c·b = d·a implies ((c, d), (a, b)) ∈ R

((a, b), (c, d)) and ((c, d), (a, b)) are symmetric.

Therefore, the relation, R, is symmetric.

Prove that R is transitive

Symbolically, transitive property is as follows; If x = y, and y = z, then x = z

From the given relation, ((a, b), (c, d)) ∈ R, then a·d = b·c

Therefore, ((c, d), (e, f)) ∈ R, then c·f = d·e

By multiplication, a·d × c·f = b·c × d·e

a·d·c·f = b·c·d·e

Therefore;

a·f·c·d = b·e·c·d

a·f = b·e

Which gives;

((a, b), (e, f)) ∈ R, therefore, the relation, R, is transitive.

Therefore;

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Based on a similar question posted online, it is required to rank the given options in the order to show that R is an equivalence relation.

Learn more about equivalent relations here:

brainly.com/question/1503196

4 0
2 years ago
10f – 6 – 10f = -8(-2f - 10) + 10
netineya [11]

Answer:

f= -6    i think

8 0
3 years ago
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