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4 years ago

Simplify the expression. 12^–3 • 12^10 • 12^0

Mathematics

1 answer:

butalik [34]4 years ago

7 0

Since they have the same base we can just add all the exponents

-3+ 10= 7

keep the base so the answer is
12^7

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valentinak56 [21]

Angle 1 and angle 5 are equal so you can set the equations equal to each other. 60 - 2x = 70 - 4x
60 + 2x = 70
2x = 10
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8 0

3 years ago

What number should be added to the expression x^2 + 6x to change it into a perfect square trinomal

alexandr402 [8]

Hello.

The correct is answer is 9.

You have to do half of 6 (3) and then square it that equals 9

So it should be <span> x^2 + 6x +9

Have a nice day</span>

5 0

3 years ago

\sqrt{12} real number whole number integer rational number

Sladkaya [172]

$\sqrt{12}$ is an irrational number

7 0

3 years ago

Write an equation in slope intercept form of a line passing through the given point (8,5) m=3

Zepler [3.9K]

If m(slope)=3, you can start to set up the equation as y=3x+b.

Using y=3x+8, you can plug in the given point being (8,5).

y=5

x=8

5=(3*8)+b

5=24+b

b=-19

The equation would be y=3x-19

3 0

3 years ago

Select the correct answer. Which expression is equivalent to 8x^2^3 sqrt 375x + 2^3 sqrt 3x^7, if x=0?

natima [27]

Answer:

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =28x^3\sqrt{3x}$

Step-by-step explanation:

The question is poorly formatted. The original question is:

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7$

We have:

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7$

Open bracket

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(8^ \frac{2}{3} *x^{3* \frac{2}{3}}) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}$

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(8^ \frac{2}{3} *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}$

Express 8 as 2^3

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(2^{3* \frac{2}{3}} *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}$

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(2^2 *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}$

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =4x^2 \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}$

Express 2^3 as 8

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{75x^3} + 8\sqrt{ 3x^7}$

Expand each exponent

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{25x^2 *3x} + 8\sqrt{ 3x * x^6}$

Split

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{25x^2} *\sqrt{3x} + 8\sqrt{3x} * \sqrt{x^6}$

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =4x^2 *5x *\sqrt{3x} + 8\sqrt{3x} * x^3$

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =20x^3\sqrt{3x} + 8x^3\sqrt{3x}$

Factorize

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =28x^3\sqrt{3x}$

4 0

3 years ago

Read 2 more answers