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emmainna [20.7K]
3 years ago
9

Find the midpoint of PQ P(4 1/3, 3 1/6), Q(-2 1/5, 3 2/3)

Mathematics
2 answers:
laiz [17]3 years ago
7 0

Answer:

(1.06, 3.41)

Step-by-step explanation:

Paha777 [63]3 years ago
6 0

Answer:

(1\frac{1}{15},3\frac{5}{12})

Step-by-step explanation:

To find the midpoint, you must average the x's then average the y's.

So the x's are 4 \frac{1}{3} and -2\frac{1}{5}.

The y's are 3\frac{1}{6} and 3\frac{2}{3}.

To find an average of a data with 2 elements you must add the two elements then take that sum and divide it by 2.

So this is what we will do with the x's then the y's.

So first x:

\frac{4\frac{1}{3}+-2\frac{1}{5}}{2}

Write as improper fractions:

\frac{\frac{13}{3}+\frac{-11}{5}}{2}

Division by 2 is the same as multiplying by 1/2:

\frac{13}{6}+\frac{-11}{10}

The least common multiple of 6 and 10 is 30.  We will multiply first fraction by 5/5 and 3/3 for the second so we can have the same denominator.

\frac{65}{30}+\frac{-33}{30}

Now that the bottoms are the same we can write as one fraction:

\frac{65+-33}{30}

Simplify:

\frac{32}{30}

If you want the answer as a mix fraction like your question began as we can do that by first figuring out how many 30's are in 32 and what is the remainder of that division.

There is one 30 and 32 and so 32-30=2 is the remainder.

1\frac{2}{30}

Reduce by dividing top and bottom by 2:

1\frac{1}{15}

Now for y:

\frac{3\frac{1}{6}+3\frac{2}{3}}{2}

Write the mix fractions as improper fractions:

\frac{\frac{19}{6}+\frac{11}{3}}{2}

Dividing by 2 is the same as multiplying by 1/2:

\frac{19}{12}+\frac{11}{6}

The least common multiple of 12 and 6 is 12 so I'm going to multiply the second fraction by 2/2 so the denominators will be the same:

\frac{19}{12}+\frac{22}{12}

Since the denominators are the same we can write as a single fraction:

\frac{41}{12}

How many 12's are in 41? 3 and so the remainder is 41-3(12)=41-36=5.

3\frac{5}{12}

So the midpoint is:

(1\frac{1}{15},3\frac{5}{12}).

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3 years ago
Which of the following represent 18=6×3
mestny [16]
C would be the correct answer

A would mean that 6=3*18, which is incorrect
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D would mean that 18=6*2 which is also incorrect

Therefore, C would be the correct answer
8 0
4 years ago
Read 2 more answers
Three weeks ago John bought stock at 49 1/4; today the stock is valued at 49 7/8. We could say the stock is performing at which
Bad White [126]

Answer:

The correct option is D.

Step-by-step explanation:

John bought stock at 49\frac{1}{4}; today the stock is valued at 49\frac{7}{8}.

49\frac{1}{4}=49+\frac{1}{4}

49\frac{1}{4}=49+0.25

49\frac{1}{4}=49.25

The value of stock at the time of purchasing is 49.25.

49\frac{7}{8}=49+\frac{7}{8}

49\frac{7}{8}=49+0.875

49\frac{7}{8}=49.875

The value of stock at current time is 49.875.

49.875-49.25=0.625

The value of stock increased by 0.625.

Since the value of stock increases, therefore the correct option is D, i.e., Above par.

7 0
4 years ago
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I-Ready<br> Which expression shows 3 2/5 + (-7 1/5) as the sum of integer and fractional parts
pochemuha
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8 0
2 years ago
Element X decays radioactively with a half life of 11 minutes. If there are 870 grams of Element X, how long, to the nearest ten
serg [7]

Answer:

It would take 27.5 minutes the element to decay to 154 grams.

Step-by-step explanation:

The decay equation:

\frac {dN}{dt}\propto -N

\Rightarrow \R\frac {dN}{dt}=-\lambda N

\Rightarrow \frac {dN}N=-\lambda  dt

Integrating both sides

\Rightarrow \int \frac {dN}N=\int-\lambda  dt

\Rightarrow ln|N|=-\lambda  t+c

When t=0, N=N_0 = initial amount

ln|N_0|=-\lambda  .0+c

\Rightarrow c=ln|N_0|

ln|N|=-\lambda t+ln|N_0|

\Rightarrow ln|N|-ln|N_0|=-\lambda t

\Rightarrow ln|\frac{N}{N_0}|=-\lambda t

Decay equation:              

                    ln|\frac{N}{N_0}|=-\lambda t

Given that, the half life of of element X is 11 minutes.

For half life, N=\frac12  N_0,  t= 11 min.

ln|\frac{N}{N_0}|=-\lambda t

\Rightarrow ln|\frac{\frac12N_0}{N_0}|=-\lambda . 11

\Rightarrow ln|\frac12}|=-\lambda . 11

\Rightarrow -\lambda . 11=ln|\frac12}|

\Rightarrow \lambda =\frac{ln|\frac12|}{-11}

\Rightarrow \lambda =\frac{ln|2|}{11}                [ ln|\frac12|=ln|1|-ln|2|=-ln|2| , since ln|1|=0]

N=154 grams, N_0 = 870 grams, t=?

ln|\frac{N}{N_0}|=-\lambda t

\Rightarrow ln|\frac{154}{870}|=-\frac{ln|2|}{11}.t

\Rightarrow t= \frac{ln|\frac{154}{870}|\times 11}{-ln|2|}

      =27.5 minutes

It would take 27.5 minutes the element to decay to 154 grams.

5 0
3 years ago
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