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dezoksy [38]
3 years ago
13

I forgot how to solve these. Please help.

Mathematics
2 answers:
AnnyKZ [126]3 years ago
6 0
This is the answer for both number 9 and number 11.
Good luck

blagie [28]3 years ago
4 0

First, lets review the shortcut rules:

multiplication means add the exponents

division means subtract the exponents

raised to a power means multiply the exponents.

negative exponents means to flip the fraction (move to the other side of the fraction bar)

9) First, I am going to distribute "-3" to "a^{-4}b²" = a¹²b^{-6}

I like to group like terms together, simplify, and then put them together:

Like terms: \frac{1}{-2} | \frac{a^{-1}}{a x a^{12} } | \frac{b^{4}}{b^{-6}}

Simplify: \frac{1}{-2} | \frac{1}{a^{14}} | \frac{b^{10}}{1}

Answer: \frac{b^{10}}{-2a^{14}}

Now try #10 on your own and check your answer below:

Answer: 2uv¹⁵



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pashok25 [27]

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Which of the following equations has an infinite number of solutions?
trasher [3.6K]

Answer:

7x + 5 = 4x + 5 + 3x

5 0
3 years ago
6x-4bracket(2x-5)&gt;8 or 5x+9&lt;2x-21
Yuri [45]

The solutions to the given inequalities are x < 6 OR x < -10

<h3>Linear Inequalities </h3>

From the question, we are to solve the given inequalities

The given inequalities are

6x - 4(2x - 5) > 8 or 5x + 9 < 2x - 21

First solve,

6x - 4(2x - 5) > 8

Clear the bracket

6x -8x + 20 > 8

6x - 8x > 8 - 20

-2x > -12

Divide both sides by -2 and flip the sign,

That is,

-2x/-2 > -12/-2

x < 6

For,

5x + 9 < 2x - 21

Subtract 2x from both sides

5x - 2x + 9 < 2x - 2x -21

3x < -21 -9

3x < -30

Divide both sides by 3

3x/3 < -30/3

x < -10

Hence, the solutions to the given inequalities are x < 6 OR x < -10

Learn more on Inequalities here: brainly.com/question/246993

#SPJ1

4 0
1 year ago
Let w = x2 + y2 + z2, x = uv, y = u cos(v), z = u sin(v). use the chain rule to find ∂w ∂u when (u, v) = (9, 0).
jasenka [17]
By the chain rule,

\dfrac{\partial w}{\partial u}=\dfrac{\partial w}{\partial x}\cdot\dfrac{\partial x}{\partial u}+\dfrac{\partial w}{\partial y}\cdot\dfrac{\partial y}{\partial u}+\dfrac{\partial w}{\partial z}\cdot\dfrac{\partial z}{\partial u}
\dfrac{\partial w}{\partial u}=2xv+2y\cos v+2z\sin v

We also have x(9,0)=0, y(9,0)=9, and z(9,0)=0, so at this point we get

\dfrac{\partial w}{\partial u}(9,0)=2\cdot9\cdot\cos0=18
7 0
3 years ago
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