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3 years ago

Answer the question about the given function.

1 answer:

5 0

F(x) = -3x^2 - 6x - 6
(-1,-3)....x = -1 and f(x) = -3 (f(x) is the same as y)

-3 = -3(-1^2) - 6(-1) - 6
-3 = -3 + 6 - 6
-3 = -3 (correct)
yes, it is a solution

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You earned 1008 in interest by keeping your money in the bank for 7 years at a rate of 4.5.What was your initial deposit?

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Answer:

The principal required to accumulate interest of $ 1,008.00 on a rate of 4.5% per year for 7 years is $ 3200.

Step-by-step explanation:

Given:

Interest I = $ 1008

Time period t = 7 years

Rate of Interest per year r = 4.5% = 4.5/100 = 0.045

To determine:

Initial deposit P = ?

Using the simple interest formula

I = Prt

P = I / rt

P = 1008 / ( 0.045 × 7 )

P = $ 3200

Therefore, the principal required to accumulate interest of $ 1,008.00 on a rate of 4.5% per year for 7 years is $ 3200.

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Step-by-step explanation:

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Solve equation for x. State any restrictions. ax+2=bx+c

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John, Sally, and Natalie would all like to save some money. John decides that it would be best to save money in a jar in his clo

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Answer:

Part 1) John’s situation is modeled by a linear equation (see the explanation)

Part 2) $y=100x+300$

Part 3) $\$12,300$

Part 4) Is a exponential growth function

Part 5) $A=6,000(1.07)^{t}$

Part 6) $\$11,802.91$

Part 7) Is a exponential growth function

Part 8) $A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

Part 9) $\$13,591.41$

Part 10) Natalie has the most money after 10 years

Step-by-step explanation:

Part 1) What type of equation models John’s situation?

Let

y ----> the total money saved in a jar

x ---> the time in months

The linear equation in slope intercept form

$y=mx+b$

The slope is equal to

$m=\$100\ per\ month$

The y-intercept or initial value is

$b=\$300$

$y=100x+300$

therefore

John’s situation is modeled by a linear equation

Part 2) Write the model equation for John’s situation

$y=100x+300$

see part 1)

Part 3) How much money will John have after 10 years?

Remember that

1 year is equal to 12 months

10 years=10(12)=120 months

For x=120 months

substitute in the linear equation

$y=100(120)+300=\$12,300$

Part 4) What type of exponential model is Sally’s situation?

we know that

The compound interest formula is equal to

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

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$P=\$6,000\\ r=7\%=0.07\\n=1$

substitute in the formula above

$A=6,000(1+\frac{0.07}{1})^{1*t}$

$A=6,000(1.07)^{t}$

therefore

Is a exponential growth function

Part 5) Write the model equation for Sally’s situation

$A=6,000(1.07)^{t}$

see the Part 4)

Part 6) How much money will Sally have after 10 years?

For t=10 years

substitute the value of t in the exponential growth function

$A=6,000(1.07)^{10}=\$11,802.91$

Part 7) What type of exponential model is Natalie’s situation?

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$P=\$5,000\\r=10\%=0.10$

substitute in the formula above

$A=5,000(e)^{0.10t}$

Applying property of exponents

$A=5,000(1.1052)^{t}$

therefore

Is a exponential growth function

Part 8) Write the model equation for Natalie’s situation

$A=5,000(e)^{0.10t}$ or $A=5,000(1.1052)^{t}$

see Part 7)

Part 9) How much money will Natalie have after 10 years?

For t=10 years

substitute

$A=5,000(e)^{0.10*10}=\$13,591.41$

Part 10) Who will have the most money after 10 years?

Compare the final investment after 10 years of John, Sally, and Natalie

Natalie has the most money after 10 years

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