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pochemuha
3 years ago
11

I say it’s D but i need reassurance

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
3 0

Answer:

Yes it's D or it's B

Step-by-step explanation:

You might be interested in
You invested $52.400 at 6% compounded annually for 5 years. What is your total return on this investment
Tems11 [23]

Answer:

$15,720

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra I</u>

Simple Interest Rate Formula: A = P(1 + rt)

  • <em>P</em> is principle amount
  • <em>r</em> is rate
  • <em>t</em> is time

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify variables.</em>

<em>P</em> = $52,400

<em>r</em> = 6%

<em>t</em> = 5

<u>Step 2: Find Current Investment Value</u>

  1. Substitute in variables [Simple Interest Rate Formula]:                                 A = 52,400(1 + 0.06 · 5)
  2. [Order of Operations] Evaluate:                                                                      A = $68,120

<u>Step 3: Find Total Return</u>

  1. Set up:                                                                                                               R = A - P
  2. Substitute in variables:                                                                                     R = 68,120 - 52,400
  3. [Order of Operations] Evaluate:                                                                      R = $15,720
5 0
3 years ago
Find the shaded area​
REY [17]

The parabola <em>y</em> = <em>x</em> ² and the line <em>x</em> + <em>y</em> = 12 intersect for

<em>x</em> ² = 12 - <em>x</em>

<em>x</em> ² + <em>x</em> - 12 = 0

(<em>x</em> - 3) (<em>x</em> + 4) = 0

===>   <em>x</em> = 3

so you can compute the area by using two integrals,

\displaystyle \int_0^3 x^2\,\mathrm dx + \int_3^{12}(12-x)\,\mathrm dx

Then the area you want is

\displaystyle \frac{x^3}3\bigg|_0^3 + \left(12x-\frac{x^2}2\right)\bigg|_3^{12} = \left(\frac{3^3}3-\frac{0^3}3\right) + \left(12^2-\frac{12^2}2 - 12\times3 + \frac{3^2}2\right) \\\\ = \boxed{\frac{99}2}

Alternatively, you can subtract the area bounded by <em>y</em> = <em>x</em> ², <em>x</em> + <em>y</em> = 12, and the <em>y</em>-axis in the first quadrant from the area of a triangle with height 12 (the <em>y</em>-intercept of the line) and length 12 (the <em>x</em>-intercept).

Such a triangle has area

1/2 × 12 × 12 = 72

and the area you want to cut away from this is given by a single integral,

\displaystyle \int_0^3 ((12-x)-x^2)\,\mathrm dx = \int_0^3(12-x-x^2)\,\mathrm dx

The integral has a value of

\displaystyle \left(12x-\frac{x^2}2-\frac{x^3}3\right)\bigg|_0^3 = 12\times3 - \frac{3^2}2 - \frac{3^3}3 \\\\ = \frac{45}2

and so the area of the shaded region is again 72 - 45/2 = 99/2.

7 0
3 years ago
Given the function f(x) = 9 - X, evaluate f (4).
Maurinko [17]
Given the function, f(x) = 9 - x :

To evaluate f(4), we must substitute 4 into the given function as an input value for x:

f(4) = 9 - 4
f(4) = 5

Therefore, f(4) = 5.
4 0
3 years ago
What goes to answer help
Oduvanchick [21]

Answer:

1/1.27

Step-by-step explanation:

6 0
3 years ago
The table of values shown below represents a linear function. Which of these points could also be an ordered pair in the table,
Amanda [17]

Answer:

b

Step-by-step explanation:

i took the quiz.

3 0
4 years ago
Read 2 more answers
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