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CaHeK987 [17]
2 years ago
5

Can someone show me how to do the problem ? If I take 8 pills a week, how many months is that?​

Mathematics
1 answer:
stiv31 [10]2 years ago
4 0

Answer:

32

Step-by-step explanation:

I assume you want to find the pills per month?  You know that you take 8 pills per week, and there are 4 weeks in a month.  So:

8 pills/week × (4 week / month) = 32 pills/month

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Which fraction is not equivalent to 7 /30
dybincka [34]
These are the ones that are not equivalent to eachother. 1. 18/30
2. 8/10
3. 4/9
4. 14/18
5. 10/20
6. 20/32
7. 18/30

8 0
3 years ago
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I BEGGG U TO HELP Me PLZZ ITS tESt<br> I WILl GiVE U BRAINLIESTTTT),:
ZanzabumX [31]

Answer:

C. 2/3

Step-by-step explanation:

5 0
3 years ago
Liam wants to buy a car. He has $100. The car that he wants costs $120,000. His parents is willing to give him $1,300. How long
mario62 [17]
Suspecting that his parents are going to give him $1,300 continually you would solve like this

$120,000-$100= $19,900
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92.23 is the amount of $1,300 donations it would take from his parents, so if he was getting $1,300 month it would take 92.23 months
5 0
2 years ago
Log(x⁴+3x³) - log(X + 3 ) + log2 - log6 = 2logx . find the value of x
AfilCa [17]

The given equation is

\begin{gathered} \log (x^4+3x^3)-\log (x+3)+\log 2-\log 6=2\log x \\ \log (\frac{x^4+3x^{3^{}}}{x+3})+\log \frac{2}{6}=\log x^2 \\ \log \frac{x^3(x+3)}{x+3}+\log \frac{1}{3}=\log x^2 \\ \log x^3+\log \frac{1}{3}=\log x^2 \\ \log \frac{x^3}{3}=\log x^2 \\ \frac{x^3}{3}=x^2 \\ x^3-3x^2=0 \\ x^2(x-3)=0 \end{gathered}

hence

x=0\text{ or x=3}

But x cannot be zero so x=3

So the value of x is 3

h

7 0
10 months ago
Two number cubes with sides numbered 1-6 are rolled. What is the probability that the sum of the numbers on the cubes is 12?
Mrrafil [7]

Answer:

1/36

Step-by-step explanation:

Attached is a chart showing the 36 possible outcomes for rolling 2 number cubes.  The sum of the two cubes is 12 in only one of the outcomes (both sixes).  The probability is 1/36.

Download pdf
6 0
2 years ago
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