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4 years ago

Keith has $400. Andrew has 3/4 as much as Keith. If Andrew gives 15% to Kevin how much does he have

1 answer:

6 0

Keith = 400. 3/4 of 400 is 300 dollars, so andrew has 300 dollars. 15% of 300 is 45 dollars, so if andrew gave keith 45 dollars, andrew would have 255 dollars left and keith would have 445.

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What is the value of X?

d1i1m1o1n [39]

Answer:

x=10

Step-by-step explanation:

We can use the angle bisector theorem to solve this problem

8 12

-------- = ---------

x+4 2x+1

Using cross products

8(2x+1) = 12(x+4)

Distribute

16x+8 = 12x+48

Subtract 12x from each side

16x -12x +8 = 12x-12x +48

4x +8 = 48

Subtract 8 from each side

4x+8-8 = 48-8

4x = 40

4x/4 = 40/4

x=10

3 0

3 years ago

Can someone give me the steps to do this like regroup help?

Ymorist [56]

The answer is 21.9

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6 0

3 years ago

Read 2 more answers

Exercise #1: The amount of money in Nicole's bank account can be represented by the function f(x) = 32.50x + 200,

kolezko [41]

The initial amount in the account was $200 (y intercept). She adds $32.50 each day to her bank account (slope).

7 0

3 years ago

Point A lies on the circle and has an x-coordinate of 1..

Neko [114]

Answer:

the last one

Step-by-step explanation:

because the distance/length formula is applied here so you just need to rearrange the coordinates in to the formula

6 0

3 years ago

Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val

Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

$A(t) = Pe^{rt}$

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that $P = 330$

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

$A(t) = Pe^{rt}$

$590 = 330e^{8r}$

$e^{8r} = \frac{590}{330}$

$e^{8r} = 1.79$

Applying ln to both sides:

$\ln{e^{8r}} = \ln{1.79}$

$8r = \ln{1.79}$

$r = \frac{\ln{1.79}}{8}$

$r = 0.0726$

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

$A(t) = 330e^{0.0726t}$

$A(10) = 330e^{0.0726*10} = 682$

The value of the account in the year 2009 will be $682.

4 0

4 years ago