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3 years ago

Simplify. Rewrite the expression in the form b^n. b^10/b^6

Mathematics

2 answers:

Vlad1618 [11]3 years ago

6 0

Answer:

<h2>The answer is b⁴</h2>

Step-by-step explanation:

$\frac{ {b}^{10} }{ {b}^{6} }$

Using the rules of indices

That's

$\frac{ {a}^{x} }{ {a}^{y} } = {a}^{x - y}$

Since the bases are the same and are dividing we subtract the exponents

So we have

$\frac{ {b}^{10} }{ {b}^{6} } = {b}^{10 - 6}$

We have the final answer as

Hope this helps you

MaRussiya [10]3 years ago

4 0

Answer:

b⁴

Step-by-step explanation:

b¹⁰ / b⁶ = b¹⁰ ⁻⁶ = b⁴

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Mothers Against Drunk Driving (MADD) is a very visible group whose main focus is to educate the public about the harm caused by

ser-zykov [4K]

Answer: 4/5

Step-by-step explanation:

--------------------------------No of vehicles involved

Did alcohol play a role - - 1 - - - - - 2 - - - - - 3

Yes---------------------------- 60 - - - - 110 - - - 30=200

No - - - - - - - - - - - - - - - - 40 - -- - 215 - - - 45 =300

- - - - - - - - - - - - - - - - - - 100 - - - - 325 - - - 75

What proportion of accidents involved more than one vehicle;

Total accident = (100 + 325 + 75) = 500

Number of accidents involving more than one vehicle = (accident involving 2 vehicles + accident involving 3 vehicles)

(325 + 75) = 400

Therefore, proportion of accidents involved more than one vehicle = (Accident involving more than one vehicle / total accidents)

= (400/ 500) = 4/5

6 0

3 years ago

4/9=w+4<br><br> solve for w.

Elina [12.6K]

Answer:

-32/9

Step-by-step explanation:

4/9=w+4

4/9-4/1=w

4-36/9=w

w=-32/9 or w=-3.56

5 0

3 years ago

Read 2 more answers

H=64t-32t^2 find the maximum height attained by the obiect

pshichka [43]

Check the picture below.

so if we just find its vertex, we know how many feet it went up by its y-coordinate.

$\bf h=64t-32t^2\implies h=-32t^2+64t+0 \\\\[-0.35em] ~\dotfill$

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h=\stackrel{\stackrel{a}{\downarrow }}{-32}t^2\stackrel{\stackrel{b}{\downarrow }}{+64}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{64}{2(-32)}~~,~~0-\cfrac{64^2}{4(-32)} \right)\implies \left( \stackrel{\stackrel{\textit{how many}}{\textit{seconds}}}{1}~~,~~\stackrel{\stackrel{\textit{how many feet}}{\textit{it went up}}}{32} \right)$