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Sphinxa [80]
3 years ago
13

How do you multiply fractions easily?

Mathematics
2 answers:
mariarad [96]3 years ago
8 0
Multiply the bottom denominators and then the same but with the numerator
love history [14]3 years ago
6 0

Answer:

To multiply two fractions, just do the following: Multiply the two numerators (top numbers) to get the numerator of the answer; multiply the two denominators (bottom numbers) to get the denominator of the answer.

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At what x value does the function given below have a hole?<br><br> f(x)=x+3/x2−9
S_A_V [24]

Answer:

hole at x=-3

Step-by-step explanation:

The hole is the discontinuity that exists after the fraction reduces. (Still doesn't exist for original of course)

The discontinuities for this expression is when the bottom is 0. x^2-9=0 when x=3 or x=-3 since squaring either and then subtracting 9 would lead to 0.

So anyways we have (x+3)/(x^2-9)

= (x+3)/((x-3)(x+3))

Now this equals 1/(x-3) with a hole at x=-3 since the x+3 factor was "cancelled" from the denominator.

4 0
3 years ago
1 2/9 + 7/24 + 23/36
damaskus [11]

Answer: 2.15277777778

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Which relationship between x and y in the equation shows a proportional relationship?
Alecsey [184]
C is the answer hope I had enjoy doing boring homework like I did

Well I actually didn't :/
7 0
3 years ago
At an interest rate of 8% compounded annually, how long will it take to double the following investments?
Paladinen [302]
Let's see, if the first one has a Principal of $50, when it doubles the accumulated amount will then be $100,

recall your logarithm rules for an exponential,

\bf \textit{Logarithm of exponentials}\\\\&#10;log_{{  a}}\left( x^{{  b}} \right)\implies {{  b}}\cdot  log_{{  a}}(x)\\\\&#10;-------------------------------\\\\&#10;\qquad \textit{Compound Interest Earned Amount}&#10;\\\\&#10;

\bf A=P\left(1+\frac{r}{n}\right)^{nt}&#10;\quad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\to &\$100\\&#10;P=\textit{original amount deposited}\to &\$50\\&#10;r=rate\to 8\%\to \frac{8}{100}\to &0.08\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{annnually, thus once}&#10;\end{array}\to &1\\&#10;t=years&#10;\end{cases}&#10;\\\\\\&#10;100=50\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 100=50(1.08)^t&#10;\\\\\\&#10;\cfrac{100}{50}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)&#10;\\\\\\&#10;

\bf log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\&#10;-------------------------------\\\\&#10;

now, for the second amount, if the Principal is 500, the accumulated amount is 1000 when doubled,

\bf \qquad \textit{Compound Interest Earned Amount}&#10;\\\\&#10;A=P\left(1+\frac{r}{n}\right)^{nt}&#10;\quad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\to &\$1000\\&#10;P=\textit{original amount deposited}\to &\$500\\&#10;r=rate\to 8\%\to \frac{8}{100}\to &0.08\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{annnually, thus once}&#10;\end{array}\to &1\\&#10;t=years&#10;\end{cases}&#10;\\\\\\&#10;1000=500\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 1000=500(1.08)^t&#10;\\\\\\&#10;

\bf \cfrac{1000}{500}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)&#10;\\\\\\&#10;log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\&#10;-------------------------------

now, for the last, Principal is 1700, amount is then 3400,

\bf \qquad \textit{Compound Interest Earned Amount}&#10;\\\\&#10;A=P\left(1+\frac{r}{n}\right)^{nt}&#10;\quad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\to &\$3400\\&#10;P=\textit{original amount deposited}\to &\$1700\\&#10;r=rate\to 8\%\to \frac{8}{100}\to &0.08\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{annnually, thus once}&#10;\end{array}\to &1\\&#10;t=years&#10;\end{cases}

\bf 3400=1700\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 3400=1700(1.08)^t&#10;\\\\\\&#10;\cfrac{3400}{1700}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)&#10;\\\\\\&#10;log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t
8 0
4 years ago
The vet says that Lena's puppy will grow to be at most 25 inches tall. Lena's puppy is currently 1 foot tall. How many more inch
mars1129 [50]

Answer:

13 inches

Step-by-step explanation:

1 foor = 12 inches

25 - 12 = 13

Therefore Lena's puppy will grow 13 more inches.

6 0
2 years ago
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