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Lerok [7]
3 years ago
14

Which linear function represents the line given by the point-slope equation y + 7 = –y plus 7 equals negative StartFraction 2 Ov

er 3 EndFraction left-parenthesis x plus 6 right-parenthesis.(x + 6)?
f(x) = –f(x) equals negative StartFraction 2 Over 3 EndFraction x minus 11.x – 11
f(x) = –f(x) equals negative StartFraction 2 Over 3 EndFraction x minus 1.x – 1
f(x) = –f(x) equals negative StartFraction 2 Over 3 EndFraction x plus 3.x + 3
f(x) = –f(x) equals negative StartFraction 2 Over 3 EndFraction x plus 13.x + 13
Mathematics
2 answers:
Citrus2011 [14]3 years ago
8 0

Answer:

Negative Start Fraction 2 Over 3 End Fraction x minus 11

f(x)=-\frac{2}{3}x-11

Step-by-step explanation:

we have

y+7=-\frac{2}{3}(x+6)

Convert to slope intercept form

y=mx+b

Solve for y

That means -----> Isolate the variable y

Distribute in the right side

y+7=-\frac{2}{3}x-\frac{2}{3}(6)

y+7=-\frac{2}{3}x-4

subtract 7 both sides

y=-\frac{2}{3}x-4-7

y=-\frac{2}{3}x-11

Convert to function notation

Let

f(x)=y

f(x)=-\frac{2}{3}x-11

therefore

The linear function is

Negative Start Fraction 2 Over 3 End Fraction x minus 11

julia-pushkina [17]3 years ago
3 0

Answer:

The sequence is recursive and can be represented by the function f(n + 1) = f(n) + f of n plus 1 equals f of n plus StartFraction 3 Over 8 EndFraction

Step-by-step explanation:

I took the test

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a. The amount that is saved at the expiration of the 5 year period is $22,769.20¢

b. The amount of interest is $2,769.20¢

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Since the amount that is deposited every year for a period of five years is $4,000 and the rate of the interest is 6.5%. We can always calculate the amount that is saved at the expiration of the five years.

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   Where P is the amount deposited per year.

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      time, t is 5 years.

   Substituting e, we have:

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a. Therefore the amount that is saved at the end of the five (5) years is $22,769.20¢

b. To find the interest, we will calculate the amount of deposit made during the period of five years and subtract the sum from the current amount that is saved ($22,769.29¢).

  Since I deposited 4,000 every year for five years, the total amount of deposit I made at the period =

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  The amount of interest is then = $22,769.20¢ - $20,000 = $2,769.20¢

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