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babunello [35]
4 years ago
13

What is the equation 5 1/3 - 2 2/3

Mathematics
1 answer:
Arada [10]4 years ago
4 0

Answer:

The answer to that equation should be about 2.67

<u><em>Hope I Helped </em></u>

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3 years ago
A website profit as a function of vistors is represented in teh table the function is quadratic
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Step-by-step explanation:

4 0
3 years ago
Simplify the expression: 2(2 + g) =
Over [174]

Answer:

4 + 2g

General Formulas and Concepts:

<u>Pre-Algebra</u>

Distributive Property

Order of Operations: BPEMDAS

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

Step-by-step explanation:

<u>Step 1: Define</u>

2(2 + g)

<u>Step 2: Expand</u>

  1. Distribute 2:                     2(2) + 2(g)
  2. Multiply:                           4 + 2g
3 0
3 years ago
Read 2 more answers
6.
Mamont248 [21]

The fixed amount is a price you pay before using equipment. So, it is the price you pay at the 0-th day: 25$.

Then, after one you pay 15 more dollars (25+15=40), and after another day you pay 15 extra dollars again (40+15=55).

So, we can see that the pattern is that you start from $25 (fixed price), and then you pay $15 per day (renting price per day)

4 0
3 years ago
Need help on problem 40 part b for integrating in respect to y! Thanks!
igomit [66]

Answer:   \bold{(a)\quad \dfrac{32}{3}\qquad (b)\quad \dfrac{32}{3}}

<u>Step-by-step explanation:</u>

(a) First, find the x-coordinates where the two equations cross

                            y = -1    and   y = 3 - x²

  -1 = 3 - x²

 -4 =     -x²

  4 =       x²

± 2 =       x       → These are the upper and lower limits of your integral

Then subtract the two equations and integrate with upper bound of x = 2 and lower bound of x = -2

<u />

\int_{-2}^{+2}[(3-x^2)-(-1)]dx\\\\\\=\int_{-2}^2(4-x^2)dx\\\\\\=4x-\dfrac{x^3}{3}\bigg|_{-2}^{+2}\\\\\\=\bigg(8-\dfrac{8}{3}\bigg)-\bigg(-8+\dfrac{8}{3}\bigg)\\\\\\=\large\boxed{\dfrac{32}{3}}

(b) We know the upper and lower bounds of the y-axis as y = 3 and y = -1

Next, find the equation that we need to integrate by solving for x.

      y = 3 - x²

x² + y = 3

x²       = 3 - y

x         =\pm\sqrt{3-y}\\

\rightarrow \qquad x=\sqrt{3-y}\quad and \quad x=-\sqrt{3-y}

Now, subtract the two equations and integrate with upper bound of y = 3 and lower bound of y = -1

\int_{-1}^{+3}[(\sqrt{3-y})-(-\sqrt{3-y})]dy\\\\\\=\int_{-1}^{+3}(2\sqrt{3-y})dy\\\\\\=\dfrac{-4\sqrt{(3-y)^3}}{3}\bigg|_{-1}^{+3}\\\\\\=\bigg(0\bigg)-\bigg(-\dfrac{32}{3}\bigg)\\\\\\=\large\boxed{\dfrac{32}{3}}

8 0
4 years ago
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