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Ber [7]
2 years ago
12

To which subsets below does the number -9/3 belong ??

Mathematics
1 answer:
emmainna [20.7K]2 years ago
4 0
Integer
because step-by-step-equation
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Simplify the exponential equation.
gavmur [86]

Answer:

b^5 * j^6

Step-by-step explanation:

Well we can use the exponential identity: x^a*x^b=x^{a+b}

The base must be the same for this to work.

So let's combine like bases: (b^2*b^3)*(j^2*j^4)

We can simplify b^2 * b^3 using this identity to get: b^(2+3) = b^5

This gives us the equation: b^5*(j^2*j^4)

But to take a deeper look as to why this identity holds, let's represent b^2 and b^3 by what it really means: (b * b) * (b * b * b), so this is really just: b * b * b * b * b which can be simplified as an exponent: b^5, hopefully this helps you understand intuitively why this identity makes sense.

So using this identity, we can simplify j^2 * j^4 to j^6

This gives us the equation: b^5 * j^6

4 0
1 year ago
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Train A & train B leave a central station at the same time. They travel at the same speed, but one in opposite directions, w
saul85 [17]

you can do it, man
I believe in you
7 0
3 years ago
Need help on this question
MariettaO [177]

<u>Given:</u>

The two equations are 2x-5y=-7 and 5x-3y=11

We need to solve the equations using elimination method.

<u>Elimination method:</u>

Let us multiply the equation 2x-5y=-7 by 5, we get;

10x-25y=-35 ---------(1)

Now, multiplying the equation 5x-3y=11 by -2, we get;

-10x+6y=-22 --------(2)

Adding equations (1) and (2), we have;

\ \ \ 10x-25y=-35\\-10x+\ \ 6y=-22\\---------\\-19y=-57

      y=3

Thus, the value of y is 3.

Substituting y=3 in the equation 2x-5y=-7, we have;

2x-5(3)=-7

  2x-15=-7

          2x=8

            x=4

Thus, the value of x is 4.

Hence, the solution of the system of equations is (4,3)

Therefore, Option A is the correct answer.

6 0
3 years ago
8.
melomori [17]

Answer:

B

Step-by-step explanation:

Communities and associative

5 0
3 years ago
The number 7 is a factor of
pantera1 [17]

Answer:

itself and numbers divisible by 7

8 0
3 years ago
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