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marshall27 [118]
3 years ago
12

Simplify Using Laws of Exponent (4x^-2)^3

Mathematics
1 answer:
lukranit [14]3 years ago
5 0

Answer:

64*x^-6

Step-by-step explanation:

Simplify Using Laws of Exponent (4x^-2)^3

Given data

(4x^-2)^3

Let us open the bracket

=4^3*x^-2*3

=64*x^-6

=64x^-6

Hence the equivalent is

64*x^-6

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Step-by-step explanation:

\dfrac{x}{-9}\geq3\qquad\text{change the signs}\\\\\dfrac{x}{9}\leq-3\qquad\text{multiply both sides by 9}\\\\9\!\!\!\!\diagup^1\cdot\dfrac{x}{9\!\!\!\!\diagup_1}\leq(-3)(9)\\\\x\leq-27

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2 years ago
Sophia's school has 23 classrooms. Each classroom has 10 tables. How many tables are there in the school? ​
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If there is 23 classrooms and each has 10 then you would have to times 23 by 10 - 23x10=230
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3 years ago
Read 2 more answers
What are the next few terms after the following sequence? 1, 1/2, 1/3, 1/4
DerKrebs [107]

Answer:

1/5, 1/6, 1/7, 1/8,1/9, 1/10,1/11,1/12

3 0
2 years ago
Need help simplifying this
diamong [38]

The simplified answer is \frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}.

<u>Step-by-step explanation:</u>

$\frac{3 y+2 x}{z+2 x}-\frac{2 y-3 x}{3 x+y}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}

To add or subtract denominators of the fraction must be same.

If it is not the same, we must take LCM of the denominators. and so we can add the fractions.

To make the denominator same multiply the 1st term (\frac{3x+y}{3x+y}) and 2nd term by (\frac{z+2x}{z+2x})

= \frac{(3 y+2 x)(3 x+y)}{(z+2 x)(3 x+y)}-\frac{(2 y-3 x)(z+2 x)}{(3 x+y)(z+2 x)}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}

LCM of the denominators is 6x²+ 3xz + 2xy +yz.

Multiply the factors in the numerator.

= \frac{\left(6 x^{2}+3 y^{2}+11 x y\right)}{(z+2 x)(3 x+y)}-\frac{\left(2 y z+4 x y-3 x z-6 x^{2}\right)}{(3 x+y)(z+2 x)}-\frac{2 z y+6 x z}{6 x^{2}+3 x z+2 x y+y z}

Now, the denominators are same, you can subtract it.

= \frac{\left(6 x^{2}+6 x^{2}+11 x y-4 x y-2 y z-2 y z+3 x z-6 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}

= \frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}

Thus the simplified solution is  \frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}

4 0
3 years ago
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