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GaryK [48]
3 years ago
5

1. Which set of points represents a function?

Mathematics
1 answer:
devlian [24]3 years ago
8 0
I think it is c but not sure
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The school choir has 84 members the ratio of girls to boys in the choir is 3:4 how many members are girls
Nostrana [21]

the answer is 36 because 84/7 = 12    12  x3 = 36
6 0
4 years ago
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
3 years ago
Read 2 more answers
A point M on a segment with endpoints X (1, −2) and Y (10, 3) partitions the segment in a 5:1 ratio. Find M. You must show all w
insens350 [35]

The coordinates of point M which divides the line segment having end points (1,-2) and (10,3) in 5:1 are (17/2,13/6).

Given that the end points of line segment  are X(1,-2) and Y(10,3) and the ratio in which the line segment is being divided is 5:1.

Line segment is a collection of points which when together joined joins two points on a surface.

The coordinates of point dividing a line segment with end points (x_{1} ,y_{1} ),(x_{2} ,y_{2} ) and ratio m:ncan be calculated using the below given formula:

(X,Y)=(mx_{2} +nx_{1} /m+n,my_{2} +ny_{1} /m+n).

We have to just put the values in the above formula to get the  coordinates.

(X,Y)=(5*10+1*1/5+1,5*3-2*1/5+1)

=(50+1/6,15-2/6)

=(51/6,13/6)

=(17/2,13/6)

Hence the coordinates of point M which divides the line segment having end points (1,-2) and (10,3) in 5:1 are (17/2,13/6).

Learn more about line segment at brainly.com/question/2437195

#SPJ1

8 0
2 years ago
How do you solve 1/4 + x = 4
kodGreya [7K]

Answer:

you subtract 1/4 from each side and get x+3 and 3/4

7 0
3 years ago
Is there enough information to prove that the triangles are congruent?
Dmitriy789 [7]

Step-by-step explanation:

The triangles are congruent due to Hypotenuse-Leg congruence. (HL congruence)

Each of the triangles have a right-angle, a congruent hypotenuse and a congruent side of the triangle.

Therefore Triangles ABC and PRQ are congruent.

3 0
3 years ago
Read 2 more answers
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