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pav-90 [236]
3 years ago
10

If 2 with a exponent of 6x =1 what is the value of X

Mathematics
2 answers:
sdas [7]3 years ago
5 0

The only exponent you can give to 2 to obtain 1 is zero:

2^z=1 \iff z=0

In fact, we have

2^z=1 \iff z = \log_2(1)=0

So, in this case, we have

2^{6x} = 1 \iff 6x = \log_2(1) \iff 6x=0\iff x=0

Lerok [7]3 years ago
3 0

the answer is

x = 0.16

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<u>ANSWER</u>

A. (4,12)

<u>EXPLANATION</u>

The equations are:

10x +2y = 64...(1)

and

3x - 4y = -36...(2)

To eliminate a variable we make the coefficients of that variable the same in both equations.

It is easier to eliminate x.

We multiply the first equation by 2 to get:

20x + 4y = 128...(3)

We add equations (2) and (3).

3x + 20x + 4y - 4y =  - 36 + 128

23x = 92

Divide both sides by 23

\frac{23x}{23}  =  \frac{92}{23}

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Put x=4 into equation (1).

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2y = 24

\frac{2y}{2}  =  \frac{24}{2}

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3 years ago
The year the Greek city of Helike disappeared is -1/5 times the year the US civil war ended. Let w represent the year the war en
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Answer:

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Answer:

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Step-by-step explanation:

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f(x)=2x^3-x^2+x-2

<u>Factor Theorem</u>

If f(a) = 0 for a polynomial then (x - a) is a factor of the polynomial f(x).

Substitute x = 1 into the function:

\implies f(1)=2(1)^3-1^2+1-2=0

Therefore, (x - 1) is a factor.

As the polynomial is cubic:

\implies  f(x)=(x-1)(ax^2+bx+c)

Expanding the brackets:

\implies  f(x)=ax^3+bx^2+cx-ax^2-bx-c

\implies  f(x)=ax^3+(b-a)x^2+(c-b)x-c

Comparing coefficients with the original polynomial:

\implies ax^3=2x^3 \implies a=2

\implies (b-a)x^2=-x^2 \implies b-2=-1 \implies b=1

\implies -c=-2 \implies c=2

Therefore:

\implies  f(x)=(x-1)(2x^2+x+2)

Cannot be factored any further.

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2 years ago
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Answer:

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