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IRINA_888 [86]
2 years ago
7

Choose one of the fractions models in Part A. Explain how to use mulitplactacion or division to check the equivalent fraction. W

hy does this work?​

Mathematics
2 answers:
Phantasy [73]2 years ago
6 0

Answer:

Find how many shaded areas there are to non shaded areas. You can use division to simplify the fraction to get your answer.

Step-by-step explanation: By dividing the fraction, you are splitting the fraction by however much you need in order to get your answer. Ex: 6/10 divided by two would be 3/5

kkurt [141]2 years ago
4 0

Answer:

It has become a cliché to describe the watch business in America

as a game of musical chairs, yet no other seems quite as

relevant.

Source: New York Times

O chord

O reinforcement

ооо

O metaphor

appendix

Step-by-step explanation:

just use ģooglr

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Help ME pls !!!!!!!!!!!!!!!
nata0808 [166]
Y>\= -40
Is the answer
6 0
2 years ago
Read 2 more answers
For the love of God help me !! I'm desperate for it tomorrow
Eduardwww [97]
Try to relax.  Your desperation has surely progressed to the point where
you're unable to think clearly, and to agonize over it any further would only
cause you more pain and frustration.
I've never seen this kind of problem before.  But I arrived here in a calm state,
having just finished my dinner and spent a few minutes rubbing my dogs, and
I believe I've been able to crack the case.

Consider this:  (2)^a negative power = (1/2)^the same power but positive.

So: 
Whatever power (2) must be raised to, in order to reach some number 'N',
the same number 'N' can be reached by raising (1/2) to the same power
but negative.

What I just said in that paragraph was:  log₂ of(N) = <em>- </em>log(base 1/2) of (N) .
I think that's the big breakthrough here.
The rest is just turning the crank.

Now let's look at the problem:

log₂(x-1) + log(base 1/2) (x-2) = log₂(x)

Subtract  log₂(x)  from each side: 

log₂(x-1) - log₂(x) + log(base 1/2) (x-2) = 0

Subtract  log(base 1/2) (x-2)  from each side:

log₂(x-1) - log₂(x)  =  - log(base 1/2) (x-2)  Notice the negative on the right.

The left side is the same as  log₂[ (x-1)/x  ]

==> The right side is the same as  +log₂(x-2)

Now you have:  log₂[ (x-1)/x  ]  =  +log₂(x-2)

And that ugly [ log to the base of 1/2 ] is gone.

Take the antilog of each side:

(x-1)/x = x-2

Multiply each side by 'x' :  x - 1 = x² - 2x

Subtract (x-1) from each side:

x² - 2x - (x-1) = 0

x² - 3x + 1 = 0

Using the quadratic equation, the solutions to that are
x = 2.618
and
x = 0.382 .

I think you have to say that <em>x=2.618</em> is the solution to the original
log problem, and 0.382 has to be discarded, because there's an
(x-2) in the original problem, and (0.382 - 2) is negative, and
there's no such thing as the log of a negative number.


There,now.  Doesn't that feel better. 
 






4 0
2 years ago
The difference of the quotient m and 9 and the quotient of n and 30
Zanzabum
Answer: (30m-9n)/270

First we start with m/9-n/30.

We multiply the denominators to get 270.

Then we multiply the numerators by the original denominators to get (30m-9n).

To check, we can use m=2, and n=4

2/9-4/30=4/45

(30*2-9*4)/270=24/270=4/45
5 0
2 years ago
Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (
tekilochka [14]

Answer:

Radius of convergence of power series is \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n        n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n       n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

Power series centered at x = a is:

\sum_{n=1}^{\infty}c_{n}(x-a)^{n}

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]

Applying the ratio test:

\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}

\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

Applying n → ∞

\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

The numerator as well denominator of \frac{a_{n}}{a_{n+1}} are polynomials of fifth degree with leading coefficients:

(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}

4 0
2 years ago
Find the value of x which satisfies the following equation.<br> log2(x−1)+log2(x+5)=4
weqwewe [10]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \: x = 3

____________________________________

\large \tt Solution  \: :

\qquad \tt \rightarrow \:  log_{2}(x - 1)  log_{2}(x + 5)  = 4

\qquad \tt \rightarrow \:  log_{2} \{(x - 1)(x + 5) \} = 4

[ log (x) + log (y) = log (xy) ]

\qquad \tt \rightarrow \: ( x - 1)(x + 5) =  {2}^{4}

\qquad \tt \rightarrow \:  {x}^{2}  + 5x - x - 5 =  16

\qquad \tt \rightarrow \:  {x}^{2}  + 4x - 5 - 16 = 0

\qquad \tt \rightarrow \:  {x}^{2}  + 4x -21 = 0

\qquad \tt \rightarrow \:  {x}^{2}  + 7x - 3x - 21 = 0

\qquad \tt \rightarrow \:  x(x + 7) - 3(x + 7) = 0

\qquad \tt \rightarrow \: (x + 7)(x - 3) = 0

\qquad \tt \rightarrow \: x =  - 7 \:  \: or \:  \: x = 3

The only possible value of x is 3, since we can't operate logarithm with a negative integer in it.

\qquad \tt \rightarrow \: x = 3

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

4 0
2 years ago
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