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tino4ka555 [31]
2 years ago
10

Multiply. 12x^4 (-5/6x^3y^2)

Mathematics
2 answers:
Darya [45]2 years ago
6 0

Answer:

-10x^{7}y^{2}

Step-by-step explanation:

12x^{4}(-\frac{5}{6} x^{3}y^{2})
First multiply 12 * - \frac{5}{6} , solving this you get -10
Next multiply x^{4} by x^{3} , solving this you get x^{7}

Finally you have y^{2}, and nothing to multiply it by. So it stays as y^{2}

Put is all together and you get -10x^{7}y^{2}

barxatty [35]2 years ago
5 0

Answer:

-10x7y2

Step-by-step explanation:

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4 0
3 years ago
Which of the following is an extraneous solution of (45 minus 3 x) Superscript one-half Baseline = x minus 9?
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The extraneous solution of the given radical equation is x = 3

<h3>Finding the extraneous solution:</h3>

Here we have the radical equation:

\sqrt{45 - 3x}  = x - 9

We want to get the extraneous solution, to get it, we first need to square both sides of the equation, so we get:

(\sqrt{45 - 3x} )^2 = (x - 9)^2\\\\45 - 3x = x^2 - 18x + 81

Now we got a quadratic equation, we can rewrite it as:

x^2 - 18x + 81 + 3x - 45 = 0\\\\x^2 - 15x + 36 = 0

To solve this we use Bhaskara's formula, we will get:

x = \frac{15 \pm \sqrt{(-15)^2 - 4*1*36} }{2*1} \\\\x = \frac{15 \pm 9}{2}

Then the two solutions are:

x = (15 + 9)/2 = 12

x = (15 - 9)/2 = 3

<h3>Which one is the extraneous solution?</h3>

Let's evaluate both solutions in our original equation and let's see which is the one that does not work:

for x  = 12 we have:

\sqrt{45 - 3*12} = 12 - 9\\3 = 3

This is true.

For x = 3 we have:

\sqrt{45 - 3*3}  = 3 - 9\\\\6 = -6

This is false, so the extraneous solution is x = 3.

If you want to learn more about extraneous solutions, you can read:

brainly.com/question/2959656

4 0
2 years ago
PLS HELP ME AND MAKE SURE TO SHOW HOW U GOT UR ANSWER
MA_775_DIABLO [31]

Answer:A.112

each piece of pizza has 28 pieces of pepporoni

so you would do 4*28 to get 112

Step-by-step explanation:

3 0
3 years ago
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