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Juliette [100K]
3 years ago
10

Write 5 to the power of 8 as a quotient of two exponential terms with the same base in four different ways. use negative or zero

exponents
Mathematics
1 answer:
ryzh [129]3 years ago
3 0
So, we are given 5^8. It was happy and content. But then... we had to write it as a quotient of two exponential terms with the same base in four different ways and use negative or zero exponents and ahhhhhh!!!

... anyways...

We'll build a quotient of two exponential terms with the same base 5. Something like this:

5^a / 5^b

We need them to make 5^8 when we are done. I'll first use a zero exponent.

[1] Now, zero exponents are nice since they make things equal 1. Like 5^0 = 1. Well, obviously, 5^8 / 1 = 5^8. So, our first quotient can be:

5^8 / 5^0

Done.

[2] Let's try this on its head. This one's a little weird. Remember that negative exponents flip things upside down. So 5^-8 = 1/5^8    and   1/5^-8 = 5^8 for example. In fact... that's the answer!

5^0 / 5^-8 = 5^8

Done.

[3] Let's try to not use 0s or 8s. We can be clever and do something like this:

5^-1 / 5^-9

What the heck is that? Well, we just flip them and get:

5^-1 / 5^-9 = 5^9 / 5^1 = 5^8

Done.

[4] Can you come up with one last trick on your own? Try it!
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Find all zeros of function and write the polynomial as a product of linear factorsX^3-x^2+9x-9
mart [117]

Answer:

\large\boxed{x=1\ \vee\ x=-3i\ \vee\ x=3i}

Step-by-step explanation:

\text{The zeros:}\\\\x^3-x^2+9x-9=0\\\\x^2(x-1)+9(x-1)=0\\\\(x-1)(x^2+9)=0\iff x-1=0\ \vee\ x^2+9=0\\\\x-1=0\qquad\text{add 1 to both sides}\\x-1+1=0+1\\\boxed{x=1}\\\\x^2+9=0\qquad\text{subtract 9 from both sides}\\x^2=-9

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3 years ago
2x+3y=1 and y=-2 - 9 solve using linear combination method. Please explain steps.
kaheart [24]

Answer:

x=17

Step-by-step explanation:

2x+3y=1

y=-2-9

  • In order to combine these two equations, an idea you need to keep in mind is finding a way of setting these equations as equal to each other. I saw that each equation shared a common value, y. In this case, we need to isolate y in the first equation so that both equations =y.

2x+3y=1

3y=-2x+1

\frac{3y}{3}=\frac{-2x+1}{3}

y=-\frac{2}{3}x+\frac{1}{3}

  • With this, we now know that both -2-9 and -\frac{2}{3}x+\frac{1}{3} are equal to y, so we can set them equal to each other.

y=-\frac{2}{3}x+\frac{1}{3}

y=-11

-\frac{2}{3}x+\frac{1}{3}=-11

  • Reply to this if anything I'm saying or doing is confusing in any way, or if you find a mistake. :) Solve for x.

-\frac{2}{3}x+\frac{1}{3}=-11

-\frac{2}{3}x=-11-\frac{1}{3}

-\frac{2}{3}x =-\frac{33}{3}-\frac{1}{3}

-\frac{2}{3}x =-\frac{34}{3}

-\frac{2}{3}x(-\frac{3}{2})=-\frac{34}{3}(-\frac{3}{2})

x=\frac{102}{6}=17

x=17

  • Hopefully this answer is correct AND makes sense in terms of how I achieved it. Again, reply to this with any questions or mistakes I made and I'll do my best to answer or fix them.

 

4 0
2 years ago
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