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Arturiano [62]
3 years ago
15

I NEED NUMBER 4 PLEEEEEASE

Mathematics
1 answer:
german3 years ago
8 0
Find the area of the first dish: since it's a square multiply the side by the side: 8•8=64. Find the area of the second dish: 9•9=81. Now subtract the first one for the second one: 81-64=17 square inches :)
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Find the slope please using the table
Verizon [17]

Answer:

11) -2/3

12) 3/4

Step-by-step explanation:

y2-y1/x2-x1

8 0
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Really easy seventh grade math can I get this problem answered? Thanks!!
gizmo_the_mogwai [7]
20/8=x/6
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5 0
3 years ago
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How many numbers are 10 units from 0 on the number line? Type your answer as a numeral.
anygoal [31]
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3 0
3 years ago
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Prove the following integration formula:
7nadin3 [17]

Answer:

See Explanation.

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Distributive Property
  • Equality Properties

<u>Algebra I</u>

  • Combining Like Terms
  • Factoring

<u>Calculus</u>

  • Derivative 1:                  \frac{d}{dx} [e^u]=u'e^u
  • Integration Constant C
  • Integral 1:                      \int {e^x} \, dx = e^x + C
  • Integral 2:                     \int {sin(x)} \, dx = -cos(x) + C
  • Integral 3:                     \int {cos(x)} \, dx = sin(x) + C
  • Integral Rule 1:             \int {cf(x)} \, dx = c \int {f(x)} \, dx
  • Integration by Parts:    \int {u} \, dv = uv - \int {v} \, du
  • [IBP] LIPET: Logs, Inverses, Polynomials, Exponents, Trig

Step-by-step Explanation:

<u>Step 1: Define Integral</u>

\int {e^{au}sin(bu)} \, du

<u>Step 2: Identify Variables Pt. 1</u>

<em>Using LIPET, we determine the variables for IBP.</em>

<em>Use Int Rules 2 + 3.</em>

u = e^{au}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv = sin(bu)du\\du = ae^{au}du \ \ \ \ \ \ \ \ \ v = \frac{-cos(bu)}{b}

<u>Step 3: Integrate Pt. 1</u>

  1. Integrate [IBP]:                                           \int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} - \int ({ae^{au} \cdot \frac{-cos(bu)}{b} }) \, du
  2. Integrate [Int Rule 1]:                                                \int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{a}{b} \int ({e^{au}cos(bu)}) \, du

<u>Step 4: Identify Variables Pt. 2</u>

<em>Using LIPET, we determine the variables for the 2nd IBP.</em>

<em>Use Int Rules 2 + 3.</em>

u = e^{au}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv = cos(bu)du\\du = ae^{au}du \ \ \ \ \ \ \ \ \ v = \frac{sin(bu)}{b}

<u>Step 5: Integrate Pt. 2</u>

  1. Integrate [IBP]:                                                  \int {e^{au}cos(bu)} \, du = \frac{e^{au}sin(bu)}{b} - \int ({ae^{au} \cdot \frac{sin(bu)}{b} }) \, du
  2. Integrate [Int Rule 1]:                                    \int {e^{au}cos(bu)} \, du = \frac{e^{au}sin(bu)}{b} - \frac{a}{b} \int ({e^{au} sin(bu)}) \, du

<u>Step 6: Integrate Pt. 3</u>

  1. Integrate [Alg - Back substitute]:     \int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{a}{b} [\frac{e^{au}sin(bu)}{b} - \frac{a}{b} \int ({e^{au} sin(bu)}) \, du]
  2. [Integral - Alg] Distribute Brackets:          \int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2} - \frac{a^2}{b^2} \int ({e^{au} sin(bu)}) \, du
  3. [Integral - Alg] Isolate Original Terms:     \int {e^{au}sin(bu)} \, du + \frac{a^2}{b^2} \int ({e^{au} sin(bu)}) \, du= \frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2}
  4. [Integral - Alg] Rewrite:                                (\frac{a^2}{b^2} +1)\int {e^{au}sin(bu)} \, du = \frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2}
  5. [Integral - Alg] Isolate Original:                                    \int {e^{au}sin(bu)} \, du = \frac{\frac{-e^{au}cos(bu)}{b} + \frac{ae^{au}sin(bu)}{b^2}}{\frac{a^2}{b^2} +1}
  6. [Integral - Alg] Rewrite Fraction:                          \int {e^{au}sin(bu)} \, du = \frac{\frac{-be^{au}cos(bu)}{b^2} + \frac{ae^{au}sin(bu)}{b^2}}{\frac{a^2}{b^2} +\frac{b^2}{b^2} }
  7. [Integral - Alg] Combine Like Terms:                          \int {e^{au}sin(bu)} \, du = \frac{\frac{ae^{au}sin(bu)-be^{au}cos(bu)}{b^2} }{\frac{a^2+b^2}{b^2} }
  8. [Integral - Alg] Divide:                                  \int {e^{au}sin(bu)} \, du = \frac{ae^{au}sin(bu) - be^{au}cos(bu)}{b^2} \cdot \frac{b^2}{a^2 + b^2}
  9. [Integral - Alg] Multiply:                               \int {e^{au}sin(bu)} \, du = \frac{1}{a^2+b^2} [ae^{au}sin(bu) - be^{au}cos(bu)]
  10. [Integral - Alg] Factor:                                 \int {e^{au}sin(bu)} \, du = \frac{e^{au}}{a^2+b^2} [asin(bu) - bcos(bu)]
  11. [Integral] Integration Constant:                     \int {e^{au}sin(bu)} \, du = \frac{e^{au}}{a^2+b^2} [asin(bu) - bcos(bu)] + C

And we have proved the integration formula!

6 0
2 years ago
Read 2 more answers
HELP PLEASE
Maurinko [17]
\bf \cfrac{x^2-x-42}{x-6}\implies \cfrac{(x+7)\underline{(x-6)}}{\underline{x-6}}\implies x+7

so.. the left-hand-side does indeed simplify to x+7, so the equation does check out.

however, notice something, for the equation of x+7, when x = 6, we get (6) + 7 which is 13.

BUT for the rational, we get    \bf \cfrac{x^2-x-42}{x-6}\qquad \boxed{x=6}\implies \cfrac{x^2-x-42}{\boxed{6}-6}\implies \stackrel{und efined}{\cfrac{x^2-x-42}{0}}

so, even though the siimplification is correct, the rational or original expression is constrained in its domain.
6 0
3 years ago
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