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3 years ago

Jeff has been offered a new job which requires that he invests $100

Mathematics

1 answer:

hichkok12 [17]3 years ago

7 0

Answer:

At the beginning of 6th month , Jeff will have in his account $ 520 .

Step-by-step explanation:

At the beginning of 6th month , Jeff will have in his account,

$ $( 100 \times (1 + 0.02\times 4) + 100 \times (1 + 0.02 \times 3) + 100 \times (1 + 0.02 \times 2) + 100 \times (1 + 0.02) + 100)$

= $ $(100 \times (5 + 0.02 \times (4 + 3 + 2 + 1)))$

= $ $(100 \times (5 + 0.02 \times 10))$

= $ $(100 \times 5.2)$

= $ 520

since, the $ 100 deposited at the end of 1st month earns interest for 4 months till the end of the 5th month, the $ 100 deposited at the end of the 2nd month earns interest for 3 months and so on till the last $100 deposited earns interest for '0' month.

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Christian purchased a new car. The value of the car is depreciating at a

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18000 represents the price of the new car.

The value of the car in 6 years can be found by substituting 6 for the variable t, as follows:

$V = 18000(1-.12)^6\\V = 18000(0.4644)\\$

In 6 years, it will be worth around $8359.27.

The equation for the car worth $20000 is this:

$V = 20000(1-.15)^6\\V = 20000(0.3771)\\$

In 6 years, the other car will be worth $7542.99, so the car worth $18000 will have more value in 6 years.

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HELP! HELP! HELP! The 12 girls at Suzie’s volleyball camp have decided to wear their camp ID’s on loops of string around their n

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Answer: No. She should have multiplied 12 and 1/2 to alive.

Step-by-step explanation:

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3 years ago

The graph of the quadratic function y = ax^2 + bx + c is given.<br> Find a + b + c.

andrey2020 [161]

Answer:

$\frac{145}{24}\approx 6$ ***

Answers may vary depending on the person reading the graph.

It is hard to tell what they think crosses nicely.

For example, I decided that the graph crossed nicely at the following points:

$(0,3)$

$(2,8)$

$(6,5)$

Step-by-step explanation:

I see that the graph crosses at $(0,3)$ , $(2,8)$ , and $(6,5)$ .

The equation $y=ax^2+bx+c$ tells us the $y$ -intercept is $c$ since when $x=0$ we have $y=a(0)^2+b(0)+c=0+0+c=c$ .

The $y$ -intercept for our graph is $3$ . Therefore, $c=3$ .

So far we have the equation:

$y=ax^2+bx+3$ .

Let's enter the other points creating a system of linear equations to solve:

$8=a(2)^2+b(2)+3$

$5=a(6)^2+b(6)+3$

Let's simplify:

$8=4a+2b+3$

$5=36a+6b+3$

Let's subtract 3 on both sides:

$5=4a+2b$

$2=36a+6b$

I choose to solve the system by elimination.

Let's multiply the top equation by -3:

$-15=-12a-6b$

$2=36a+6b$

Now adding the equations results in:

$-13=24a$

Divide both sides by 24:

$\frac{-13}{24}=a$

Now using one of the equations we can find $b$ :

$5=4a+2b$ with $a=\frac{-13}{24}$

$5=4\frac{-13}{24}+2b$

$5=\frac{-13}{6}+2b$

Add 13/6 on both sides:

$5+\frac{13}{6}=2b$

$\frac{30+13}{6}=2b$

$\frac{43}{6}=2b$

Divide both sides by 2:

$\frac{43}{2(6)}=b$

$\frac{43}{12}=b$

So we have the equation:

$y=\frac{-13}{24}x^2+\frac{43}{12}+3$

Let's evaluate $a+b+c$ now:

$\frac{-13}{24}+\frac{43}{12}+3$

$\frac{-13+86+72}{24}$

$\frac{73+72}{24}$

$\frac{145}{24}$

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3 years ago

If you need $3,000 four years from now, how much would you need to deposit into an account that earns 5% interest, compounded an

Aleks04 [339]

Answer:

2468.11 dollars

Step-by-step explanation:

Amount that should be available after four years =3000 $

Interest rate = 5%

No of years =4

Let P be the principal invested

Then $3000=P(1+0.05)^4\\P=3000(1.05)^{-4} \\=2468.11$

Hence 2468.11 dollars should be deposited to get 3000 dollars after 4 years at 5% interest.

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2 years ago

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Step-by-step explanation:

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3 years ago

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