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Cloud [144]
3 years ago
14

When doing division and writing down the answer does the "remainder" get put behind a decimal point? (18 POINTS)

Mathematics
1 answer:
torisob [31]3 years ago
5 0
Yes 4.5 would be 4 wholes and .5 or 1/2 away from making another whole
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Select all the values that are equivalent to the given expression. Express your answer in scientific notation.
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the first and last ones

Step-by-step explanation:

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A certain drug is made from only two ingredients: compound A and compound B. There are 3 milliliters of compound A used for ever
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375 milliliters of compound B

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I honestly have no clue on this one, does anyone have a clue?! Please!
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Due to delays, a company had to extend a project from 109 days to 123 days. What was the percentage increase in time to complete
Bumek [7]

Answer:

1% or 1.1%

                     I'm 60% sure

Step-by-step explanation:

So what i did was divide 123 by 109

6 0
3 years ago
Joe's annual income has been increasing in the same dollar amount. The first year his income was $15,200, and the 4th year his i
lidiya [134]

Answer:

Therefore 6th year his income was $19,700.

Step-by-step explanation:

Given, Joe's annual income has been increasing in some dollar amount . The first year his income was $15,200 and 4th  year his income was $17,900.

A=$17900, P= $15,200 and n = 3 year

A= P(1+\frac{r}{100} )^n

\Leftrightarrow 17900=15200(1+\frac{r}{100})^3

\Leftrightarrow (1+\frac{r}{100})^3=\frac{17900}{15200}

\Leftrightarrow (1+\frac{r}{100})=(\frac{17900}{15200})^{\frac{1}{3} }

\Leftrightarrow \frac{r}{100}=(\frac{17900}{15200})^{\frac{1}{3} }-1

\Leftrightarrow r=5.60

Let  t^{th} year Joe's income was $19,700.

\therefore 19700=15,200(1+\frac{5.60}{100} )^{t-1}

\Leftrightarrow 1.296= (1+0.056)^{t-1}

\Leftrightarrow 1.296= (1.056)^{t-1}

\Leftrightarrow  log(1.296)= (t-1)log(1.056)

\Leftrightarrow  \frac{log(1.296)}{log(1.056)}= (t-1)

\Leftrightarrow t-1= 4.75

\Leftrightarrow t= 4.75+1

\Leftrightarrow t = 5.75 ≈6

Therefore 6th year his income was $19,700.

8 0
3 years ago
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