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2 years ago

34 X 19/38 reduced to lowest term

Mathematics

1 answer:

torisob [31]2 years ago

7 0

19/38 = 1/2 (divided top and bottom by 19)

34 X 19/38

= 34 X 1/2

= 34/2

= 17/1

Answer

17/1 (or 17)

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guapka [62]

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Answer:

d = 3√10 ≈ 9.487

Step-by-step explanation:

The distance formula is appropriate.

d = √((x2 -x1)^2 +(y2 -y1)^2)

d = √((-3 -0)^2 +(2 -(-7))^2) = √(9 +81) = √90

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2 years ago

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2 years ago

Show work please help

Snezhnost [94]

Answer:

20 yd^2

Step-by-step explanation:

Your work is partially correct.

Assuming that the sides marked 8 yds and 2 yds are parallel, then the area of the trapezoid is

A = ( 8 yds + 2 yds)

------------------------ * 4 = 20 yd^2

4 0

2 years ago

When the effective interest rate is 9% per annum, what is the present value of a series of 50 annual payments that start at $100

ser-zykov [4K]

Answer:

$1,109.62

Step-by-step explanation:

Let's first compute the future value FV.

In order to see the rule of formation, let's see the value (in $) for the first few years

End of year 0

1,000

End of year 1(capital + interest + new deposit)

1,000*(1.09)+10

End of year 2 (capital + interest + new deposit)

(1,000*(1.09)+10)*1.09 +10 =

$\bf 1,000*(1.09)^2+10(1+1.09)$

End of year 3 (capital + interest + new deposit)

$\bf (1,000*(1.09)^2+10(1+1.09))(1.09)+10=\\1,000*(1.09)^3+10(1+1.09+1.09^2)$

and we can see that at the end of year 50, the future value is

$\bf FV=1,000*(1.09)^{50}+10(1+1.09+(1.09)^2+...+(1.09)^{49}$

The sum

$\bf 1+1.09+(1.09)^2+...+(1.09)^{49}$

is the sum of a geometric sequence with common ratio 1.09 and is equal to

$\bf \frac{(1.09)^{50}-1}{1.09-1}=815.08356$

and the future value is then

$\bf FV=1,000*(1.09)^{50}+10*815.08356=82,508.35564$

The present value PV is

$\bf PV=\frac{FV}{(1.09)^{50}}=\frac{82508.35564}{74.35572}=1,109.616829\approx \$1,109.62$

rounded to the nearest hundredth.

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Step-by-step explanation:

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