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Andrews [41]
2 years ago
8

Ben deposited $6,500 in a simple interest account that pays 2.8% interest annually. If Ben leaves the money in the account for 1

2 years, how much interest will he earn?

Mathematics
1 answer:
svet-max [94.6K]2 years ago
5 0

Answer:

$2,184

Step-by-step explanation:

$6,500 x 2.8% = annual interest

6,500 x 2.8% = $182 annual interest

$182 x 12 years = interest earned in 12 years.

182 x 12 = $2,184

You might be interested in
Solve the following quadratics. State the FACTORS AND SOLUTIONS. 1. 2x^2 - 7x + 3 2. 3x^2 + 7x +2
tekilochka [14]

Answer:

1. x = 3, 1/2 (solutions); (x - 3)(2x - 1) (factors)

2. x = -1/3, -2 (solutions); (3x + 1)(x + 2) (factors)

Step-by-step explanation:

<u>1. 2x^2 - 7x + 3</u>

To solve problem 1, you will need to identify your a, b, and c values in this quadratic function.

Since this problem is in standard form, it will be easy to identify these values. The standard form of a quadratic function is ax^2 + bx + c.

The a value is 2, the b value is -7, and the c value is 3 if we use our standard form and see which numbers are plugged into it.

Since we know that

  • a = 2
  • b = -7
  • c = 3

we can use the quadratic formula: x = \frac{-b~\pm~\sqrt{b^2~-~4ac} }{2a}

Substitute the a, b, and c values into the quadratic formula: x=\frac{-(-7)\pm\sqrt{(-7)^2-4(2)(3)} }{2(2)}

Now simplify using the laws of pemdas: x=\frac{7\pm\sqrt{(49)-(24)} }{4}

Simplify even further: x=\frac{7\pm\sqrt{(25)} }{4} \rightarrow x=\frac{7\pm (5) }{4}

Now split this equation into two equations to solve for x: x=\frac{12 }{4} ~~and~~ x=\frac{2 }{4}

12/4 can be simplified to 3, and 2/4 can be simplified to 1/2.

This means your solutions to problem 1 is 3, 1/2.

\boxed {x=3,\frac{1}{2} }

There is also another way to solve for the quadratic functions, and this was by factoring.

If you factor 2x^2 - 7x + 3 using the bottoms-up method, you will get (x - 3)(2x - 1).

After factoring, solving for the solutions is simple because all you have to do is set each factor to 0.

  • x - 3 = 0
  • 2x - 1 = 0

After solving for x by adding 3 to both sides, or by adding 1 to both sides then dividing by 2, you will end up with the same solutions: x = 3 and x = 1/2.

<u>2. 3x^2 + 7x + 2</u>

To save time I'll be using the bottoms-up factoring method, but remember to refer back to problem 1 (quadratic formula) if you prefer that method.

Factor this quadratic function using the bottoms-up method. After factoring you will have (3x + 1)(x + 2). These are your factors.

Now to solve for x and find the solutions of the quadratic function, you will set both factors equal to 0.

  • 3x + 1 = 0
  • x + 2 = 0

Solve.

<u>First factor:</u> 3x + 1 = 0

Subtract 1 from both sides.

3x = -1

Divide both sides by 3.

x = -1/3

<u>Second factor:</u> x + 2 = 0

Subtract 2 from both sides.

x = -2

Your solutions are x = -1/3 and x = -2.

\boxed {x = -\frac{1}{3} , -2}

7 0
3 years ago
I need help thank you
Juliette [100K]

Answer:


Step-by-step explanation:

The key and the most critical step is to draw the line first. Don't do anything else before that.

(2,-8) (-2,-8)

What you find out is that the y value is always - 8.

So the line has no slope and any x value

y = 0*x + b

y = - 8 is the equation of the line.

8 0
3 years ago
Read 2 more answers
Which equation is true for the value b = 2?
marta [7]
A. b = 3 
B. b = 2 
C. b = 4 
D. b = 3/2 

So, here I state the answer to this problem is (B) 

Have a great day. 
Glad to help out! 
5 0
3 years ago
1.) Divide $80 among three people so that the second will have twice as much as the first, and the third will have $5 less than
Marina CMI [18]
1 ) The first person will have $17. The second person will have $34 and the third person will $29.
5 0
3 years ago
140 subtracted from the square of a number the result is three times the number find the negative solution
natima [27]

Answer:

Suppose~that~the~number~is~x.\\According~to~question,\\x^2-140=3x\\or, x^2-3x-140=0\\From~the~quadratic~formula,\\x = \frac{-(-3)+\sqrt{(-3)^2-4(1)(-140)} }{2(1)} ~and ~x= \frac{-(-3)-\sqrt{(-3)^2-4(1)(-140)} }{2(1)} \\or, x = \frac{3+\sqrt{9+560} }{2}~and~x=\frac{3-\sqrt{9+569} }{2}\\or, x = \frac{3+\sqrt{569} }{2}~and~x=\frac{3-\sqrt{569} }{2}\\So,~the~required~negative~number~is~\frac{3-\sqrt{569} }{2}.

5 0
3 years ago
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