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Andrew [12]
3 years ago
8

Latasha can read 30 pages of economics in an hour. She can also read 20 pages of sociology in an hour. She spends 4 hours per da

y studying. Her opportunity cost of reading 10 pages of sociology is ___ pages of economics. Her opportunity cost of reading 18 pages of economics is ____ pages of sociology. Enter whole numbers.
Mathematics
1 answer:
stiv31 [10]3 years ago
8 0

Answer:

Her opportunity cost of reading 10 pages of sociology is 15 pages of Economics

Her opportunity cost of reading 18 pages of economics is 12 pages of sociology.

Step-by-step explanation:

For economics, Latasha reads 1 page in 2 mins (gotten by dividing the number of pages by the time taken to read it i.e 30 pages in 1 hr or 60 mins)

Her sociology is read 1 page in 3 mins (20 pages read in 60 mins)

To read 10 pages of sociology will take 30 mins. She will read 15 pages of Economics in the same time which is the opportunity cost.

In the same vein, It will take her 36 mins to read 18 pages of Economics at the same rate. This is the opportunity cost of reading 12 pages of Sociology (which she reads at 1 page in 3 mins).

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Answer: B) A = 750(1.04)ⁿ

<u>Step-by-step explanation:</u>

The formula for compounded annually is: A = P(1 + r)ⁿ   where

  • A (amount accrued) = <em>unknown</em>
  • P (amount invested) = $750
  • r (interest rate) = 4% -->(0.04)
  • t (time in years) = <em>unknown</em>

A = 750(1 + 0.04)ⁿ

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3 years ago
HELP PLEASE!!! Which of the following is the equation for X: ax^2 = bx? Select all that apply.
ki77a [65]
Ax^2 = bx Divide by x
ax^2/x = b
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3 years ago
What is 3n + 17 = 44??
sveta [45]

Answer:

n=9

Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM​
Snezhnost [94]

\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}

  • Given - <u>A </u><u>trapezium</u><u> </u><u>ABCD </u><u>with </u><u>non </u><u>parallel </u><u>sides </u><u>of </u><u>measure </u><u>1</u><u>5</u><u> </u><u>cm </u><u>each </u><u>!</u><u> </u><u>along </u><u>,</u><u> </u><u>the </u><u>parallel </u><u>sides </u><u>are </u><u>of </u><u>measure </u><u>1</u><u>3</u><u> </u><u>cm </u><u>and </u><u>2</u><u>5</u><u> </u><u>cm</u>

  • To find - <u>Area </u><u>of </u><u>trapezium</u>

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

<u>Construction</u><u> </u><u>-</u>

draw \: CE \: \parallel \: AD \:  \\ and \: CD \: \perp \: AE

Now , we can clearly see that AECD is a parallelogram !

\therefore AE = CD = 13 cm

Now ,

AB = AE + BE \\\implies \: BE =AB -  AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm

Now , In ∆ BCE ,

semi \: perimeter \: (s) =  \frac{15 + 15 + 12}{2}  \\  \\ \implies \: s =  \frac{42}{2}  = 21 \: cm

Now , by Heron's formula

area \: of \: \triangle \: BCE =  \sqrt{s(s - a)(s - b)(s - c)}  \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)}  \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21}  \: cm {}^{2}

Also ,

area \: of \: \triangle \:  =  \frac{1}{2}  \times base \times height \\  \\\implies 18 \sqrt{21} =  \: \frac{1}{\cancel2}  \times \cancel12  \times height \\  \\ \implies \: 18 \sqrt{21}  = 6 \times height \\  \\ \implies \: height =  \frac{\cancel{18} \sqrt{21} }{ \cancel 6}  \\  \\ \implies \: height = 3 \sqrt{21}  \: cm {}^{2}

<u>Since </u><u>we've </u><u>obtained </u><u>the </u><u>height </u><u>now </u><u>,</u><u> </u><u>we </u><u>can </u><u>easily </u><u>find </u><u>out </u><u>the </u><u>area </u><u>of </u><u>trapezium </u><u>!</u>

Area \: of \: trapezium =  \frac{1}{2}  \times(sum \: of \:parallel \: sides) \times height \\  \\ \implies \:  \frac{1}{2}  \times (25 + 13) \times 3 \sqrt{21}  \\  \\ \implies \:  \frac{1}{\cancel2}  \times \cancel{38 }\times 3 \sqrt{21}  \\  \\ \implies \: 19 \times 3 \sqrt{21}  \: cm {}^{2}  \\  \\ \implies \: 57 \sqrt{21}  \: cm {}^{2}

hope helpful :D

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What is the value of x?​
Assoli18 [71]
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