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frosja888 [35]
3 years ago
10

What is the value of the expression 3/3^-2 ?

Mathematics
1 answer:
kotykmax [81]3 years ago
4 0
The answer is a I hope this helped you out
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The table and graph below record the data for the depth of the tide. use the information to identify the frequency of the graph
scoundrel [369]

Answer:

This is to complex

Step-by-step explanation: This is more complex I am sure even your teacher needs the answers from a book to answer this not gonna cap

8 0
3 years ago
Find the area of the region enclosed by the graphs of these equations. (CALCULUS HELP)
sergiy2304 [10]

Answer:

\displaystyle A = \frac{20\sqrt{15}}{3}

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

Equality Properties

  1. Multiplication Property of Equality
  2. Division Property of Equality
  3. Addition Property of Equality
  4. Subtraction Property of Equality

<u>Algebra I</u>

  • Terms/Coefficients
  • Graphing
  • Exponential Rule [Root Rewrite]:                                                                   \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Area - Integrals

U-Substitution

Integration Rule [Reverse Power Rule]:                                                               \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

Integration Property [Multiplied Constant]:                                                         \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

Integration Property [Addition/Subtraction]:                                                       \displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx

Integration Rule [Fundamental Theorem of Calculus 1]:                                     \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

Area of a Region Formula:                                                                                     \displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx

Step-by-step explanation:

<u>Step 1: Define</u>

F: y = √(15 - x)

G: y = √(15 - 3x)

H: y = 0

<u>Step 2: Find Bounds of Integration</u>

<em>Solve each equation for the x-value for our bounds of integration.</em>

F

  1. Set <em>y</em> = 0:                                                                                                         0 = √(15 - x)
  2. [Equality Property] Square both sides:                                                          0 = 15 - x
  3. [Subtraction Property of Equality] Isolate <em>x</em> term:                                         -x = -15
  4. [Division Property of Equality] Isolate <em>x</em>:                                                        x = 15

G

  1. Set y = 0:                                                                                                         0 = √(15 - 3x)
  2. [Equality Property] Square both sides:                                                          0 = 15 - 3x
  3. [Subtraction Property of Equality] Isolate <em>x</em> term:                                         -3x = -15
  4. [Division Property of Equality] Isolate <em>x</em>:                                                        x = 5

This tells us that our bounds of integration for function F is from 0 to 15 and our bounds of integration for function G is 0 to 5.

We see that we need to subtract function G from function F to get our area of the region (See attachment graph for visual).

<u>Step 3: Find Area of Region</u>

<em>Integration Part 1</em>

  1. Rewrite Area of Region Formula [Integration Property - Subtraction]:     \displaystyle A = \int\limits^b_a {f(x)} \, dx - \int\limits^d_c {g(x)} \, dx
  2. [Integral] Substitute in variables and limits [Area of Region Formula]:     \displaystyle A = \int\limits^{15}_0 {\sqrt{15 - x}} \, dx - \int\limits^5_0 {\sqrt{15 - 3x}} \, dx
  3. [Area] [Integral] Rewrite [Exponential Rule - Root Rewrite]:                       \displaystyle A = \int\limits^{15}_0 {(15 - x)^{\frac{1}{2}}} \, dx - \int\limits^5_0 {(15 - 3x)^{\frac{1}{2}}} \, dx

<u>Step 4: Identify Variables</u>

<em>Set variables for u-substitution for both integrals.</em>

Integral 1:

u = 15 - x

du = -dx

Integral 2:

z = 15 - 3x

dz = -3dx

<u>Step 5: Find Area of Region</u>

<em>Integration Part 2</em>

  1. [Area] Rewrite [Integration Property - Multiplied Constant]:                       \displaystyle A = -\int\limits^{15}_0 {-(15 - x)^{\frac{1}{2}}} \, dx + \frac{1}{3}\int\limits^5_0 {-3(15 - 3x)^{\frac{1}{2}}} \, dx
  2. [Area] U-Substitution:                                                                                   \displaystyle A = -\int\limits^0_{15} {u^{\frac{1}{2}}} \, du + \frac{1}{3}\int\limits^0_{15} {z^{\frac{1}{2}}} \, dz
  3. [Area] Reverse Power Rule:                                                                         \displaystyle A = -(\frac{2u^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15} + \frac{1}{3}(\frac{2z^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15}
  4. [Area] Evaluate [Integration Rule - FTC 1]:                                                   \displaystyle A = -(-10\sqrt{15}) + \frac{1}{3}(-10\sqrt{15})
  5. [Area] Multiply:                                                                                               \displaystyle A = 10\sqrt{15} + \frac{-10\sqrt{15}}{3}
  6. [Area] Add:                                                                                                     \displaystyle A = \frac{20\sqrt{15}}{3}

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Area Under the Curve - Area of a Region (Integration)

Book: College Calculus 10e

3 0
3 years ago
Plzz HELPP I DONT GOT TIME LOL
GaryK [48]
Radius = circumference / (2 * PI)
radius = 40 / (2 * PI)
<span><span>radius = 6.3661977237 </span> inches
diameter = </span> <span> <span> <span> 12.7323954474 </span> </span> </span>
diameter = 12.73 inches
The 14 inch pizza tray is larger than 12.73 inches and so the pizza will fit.


5 0
3 years ago
Solve fast please. its related to mathematics...​
nataly862011 [7]

Answer:

The answer is 6

Step-by-step explanation:

\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16} \\\\= \lim_{x \to 4} \frac{x^3 - 64}{(x + 4)(x - 4)} \\\\=\lim_{x \to 4} \frac{(x^2 + 4x + 16)(x - 4)}{(x + 4)(x - 4)} \\\\= \lim_{x \to 4} \frac{(x^2 + 4x + 16)}{(x + 4)} \\\\= (16 + 16 + 16) / 8\\= 48 / 8\\= 6

8 0
3 years ago
Write the equation of a line that passes through (4,-10) and has a slope of 2, please!
Alex Ar [27]

Answer:

y = 2x - 18

Step-by-step explanation:

There are lots of different methods to do these types of questions, but this is the way i like to do it.

Coordinate (x1, y1) : (4, -10) and Gradient (m): 2

Start with this generic equation:

y- y1 = m(x-x1)

Substitute in the values they have given you:

y - (-10) = 2(x-4)

This becomes:

y + 10 = 2x - 8

Keep solving by moving the 10 over to the other side:

y = 2x -8-10

y = 2x - 18

<u>An easy way to double check:</u>

1) the coefficient of x is the slope. We got 2x, and the slope is 2, so that's correct.

2) Substitute x=4 into the equation, and we get -10, which is the exact coordinate that they gave us. This is also right.

Hope this helped : )

5 0
4 years ago
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