All categories

2 years ago

Simplify 8^-5 over 8^-3

Mathematics

2 answers:

Alex2 years ago

7 0

Answer:

64-5 divided by 64-3 is 59 divided by 61. 59

8 squared is equal to 64. -----

Step-by-step explanation:

trasher [3.6K]2 years ago

7 0

$\bf ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{8^{-5}}{8^{-3}}\implies \cfrac{1}{8^{-3}\cdot 8^5}\implies \cfrac{1}{8^{-3+5}}\implies \cfrac{1}{8^2}\implies \cfrac{1}{64}$

You might be interested in

If r(x) = 2 – x2 and w(x) = x – 2, what is the range of (w circle r) (x)?

VashaNatasha [74]

Answer:

$(-\infty, 0)$

Step-by-step explanation:

(w circle r) (x) is the composite function(w of r(x)), that is, w(r(x))[/tex]

We have that:

$r(x) = 2 - x^{2}$

$w(x) = x - 2$

Composite function:

$w(r(x)) = w(2 - x^{2}} = 2 - x^{2} - 2 = -x^{2}$

$-x^{2}$ is a negative parabola with vertex at the original.

So the range(the values that y assumes), is:

$(-\infty, 0)$

3 0

2 years ago

Read 2 more answers

I need help with this answer

Llana [10]

Answer:

Im not sure but lets play warzone

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Tamiko makes a necklace by stringing beads with the following colors: red, yellow, blue, red, yellow, blue,... If she continues

Anettt [7]

$50=3\cdot16+2$

So, it will be the second color which is yellow.

6 0

2 years ago

Read 2 more answers

What’s the answer pls help me

defon

Answer:

H Maggie is write answers

8 0

2 years ago

Would appreciate the help !

aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

$\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 2

$\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity $\sin^2(\theta)+\cos^2(\theta) = 1$ in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

3 0

2 years ago