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Brut [27]
3 years ago
14

What is the present value of 6000 received 14 years from today when the interest rate is 10%

Mathematics
1 answer:
likoan [24]3 years ago
3 0

Answer:

$14400

Step-by-step explanation:

Given data

Principal= 6000

Time= 14 years

Rate= 10%

Let us use the simple interest expression to find the final amount

A=P(1+rt)

substitute

A=6000(1+0.1*14)

A=6000(1+1.4)

A=6000*2.4

A=$14400

Hence based on a simple interest model the final amount is $14400

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3 years ago
Solve ax-b/d=c, for x
Kitty [74]

Answer:

x = cd + b / a

Step-by-step explanation:

ax - b / d = c

ax - b = cd

ax = cd + b

x = cd + b / a

5 0
3 years ago
What is the equation of the function shown in the graph provided?<br><br> y= _ x + _
Mkey [24]

Answer:

y = -3x - 6

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS

<u>Algebra I</u>

Slope Formula: m=\frac{y_2-y_1}{x_2-x_1}

Slope-Intercept Form: y = mx + b

  • m - slope
  • b - y-intercept

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Find points from graph.</em>

x-intercept (-2, 0)

y-intercept (0, -6)

<u>Step 2: Find slope </u><em><u>m</u></em>

  1. Substitute:                  m=\frac{-6-0}{0+2}
  2. Subtract/Add:             m=\frac{-6}{2}
  3. Divide:                        m=-3

<u>Step 3: Redefine</u>

Slope <em>m</em> = -3

y-intercept <em>b</em> = -6

<u>Step 4: Write linear equation</u>

Slope-Intercept Form:   y = -3x - 6

7 0
3 years ago
If g is the inverse of the function f(x) = sqrt(x - 6) - 10 , which of the following is g?
iren2701 [21]

Answer:

  • g(x) = x² + 20x + 106

Step-by-step explanation:

<u>Given:</u>

  • f(x) =\sqrt{x-6}-10

If g(x) is the inverse of f(x), find it.

<u>Swap x with g(x) and f(x) with x:</u>

  • x =\sqrt{g(x)-6}-10

<u>Solve for g(x):</u>

  • x + 10 = \sqrt{g(x)-6}
  • (x+10)^2=g(x)-6
  • g(x) = x^2+20x + 100+6
  • g(x) = x^2+20x + 106

3 0
2 years ago
Please help me quickly
lys-0071 [83]
For question 6 the answer is 100 because a quarter of a circle is 90 degrees
4 0
2 years ago
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