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Brilliant_brown [7]
3 years ago
10

Which decimal is equivalent to 8.100

Mathematics
1 answer:
larisa86 [58]3 years ago
7 0

Answer:

81/10

Step-by-step explanation:

i just figured it out i really don't have one but i hope it helps

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Evaluate each expression for n = 5.<br> 1. 9n - 9<br> 2. (3n + 5n)10<br> 3. 12 x n/2
erma4kov [3.2K]

Step-by-step explanation:

Given

n = 5

1. 9n - 9

= 9 * 5 - 9

= 45 - 9

= 36

2 . (3n + 5n) 10

= ( 3 * 5 + 5* 5)10

= 40* 10

= 400

3 . 12 * n/2

= 12 * 5 / 2

= 6 * 5

= 30

Hope it will help :)

4 0
3 years ago
Find both the number of combinations and the number of permutations for the given number of objects.
Nikitich [7]

Answer:

First, if we have a set of K elements, such that are ordered as:

{x₁, x₂, ...}

The total number of permutations for the K elements can be found in the next way.

For the first element in the set, we have K options.

For the second element in the set, we have (K - 1) options (because we already choose one)

For the third element we have (K - 2) options, and so on.

The total number of permutation is equal to the product between the numbers of options for each position's element, then the number of permutations for K elements is:

permutations = K*(K - 1)*(K - 2)*....*2*1 = K!

Now suppose that we have a set of N elements, and we want to make groups of K elements.

The total number of different combinations of K elements is given by the equation:

C(N, K) = \frac{N!}{(N - K)!*K!}

In this case we have 15 objects (then  N = 15) and we take 7 at the time (Then K = 7)

Where we need to take in account the number of combinations and also the permutations for each combination.

Then the total number of different sets is:

C(15*7)*7!

First, the total number of combinations will be:

C(15,7) = \frac{15!}{(15 - 7)!7!} = \frac{15!}{8!*7!}  = \frac{15*14*13*12*11*10*9}{7*6*5*4*3*2*1}  = 6,435

So we have 6,436 combinations, and each one of these combinations has 7! permutations.

permutations = 7! = 7*6*5*4*3*2*1 = 5,040

if we combine these we get:

Combinations*Permutations = 6,435*5,040 = 32,432,400

5 0
3 years ago
What’s the volume of one cylinder?
r-ruslan [8.4K]

Answer:

For the first one its: Volume = 402.12in³

For the second one its: Volume = 201.0619 in³

Step-by-step explanation:

For the first one work:

Volume = 3.1416 x 42 x 8

= 3.1416 x 16 x 8

For the second one work:

Volume = 3.1416 x 42 x 4

= 3.1416 x 16 x 4

5 0
3 years ago
Read 2 more answers
Cat e (5×4)+(5×7) ? plss​
KengaRu [80]

Answer:

55

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Who can help me d e f thanks​
12345 [234]

d)

y = (2ax^2 + c)^2 (bx^2 - cx)^{-1}

Product rule:

y' = \bigg((2ax^2+c)^2\bigg)' (bx^2-cx)^{-1} + (2ax^2+c)^2 \bigg((bx^2-cx)^{-1}\bigg)'

Chain and power rules:

y' = 2(2ax^2+c)\bigg(2ax^2+c\bigg)' (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} \bigg(bx^2-cx\bigg)'

Power rule:

y' = 2(2ax^2+c)(4ax) (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} (2bx - c)

Now simplify.

y' = \dfrac{8ax (2ax^2+c)}{bx^2 - cx} - \dfrac{(2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}

y' = \dfrac{8ax (2ax^2+c) (bx^2 - cx) - (2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}

e)

y = \dfrac{3bx + ac}{\sqrt{ax}}

Quotient rule:

y' = \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{\left(\sqrt{ax}\right)^2}

y'= \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{ax}

Power rule:

y' = \dfrac{3b \sqrt{ax} - (3bx+ac) \left(-\frac12 \sqrt a \, x^{-1/2}\right)}{ax}

Now simplify.

y' = \dfrac{3b \sqrt a \, x^{1/2} + \frac{\sqrt a}2 (3bx+ac) x^{-1/2}}{ax}

y' = \dfrac{6bx + 3bx+ac}{2\sqrt a\, x^{3/2}}

y' = \dfrac{9bx+ac}{2\sqrt a\, x^{3/2}}

f)

y = \sin^2(ax+b)

Chain rule:

y' = 2 \sin(ax+b) \bigg(\sin(ax+b)\bigg)'

y' = 2 \sin(ax+b) \cos(ax+b) \bigg(ax+b\bigg)'

y' = 2a \sin(ax+b) \cos(ax+b)

We can further simplify this to

y' = a \sin(2(ax+b))

using the double angle identity for sine.

7 0
2 years ago
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