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](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Be%5Eu%5D%3Du%27e%5Eu)
- Integration Constant C
- Integral 1:

- Integral 2:

- Integral 3:

- Integral Rule 1:

- Integration by Parts:

- [IBP] LIPET: Logs, Inverses, Polynomials, Exponents, Trig
Step-by-step Explanation:
<u>Step 1: Define Integral</u>

<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>Step 3: Integrate Pt. 1</u>
- Integrate [IBP]:

- Integrate [Int Rule 1]:

<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>Step 5: Integrate Pt. 2</u>
- Integrate [IBP]:

- Integrate [Int Rule 1]:

<u>Step 6: Integrate Pt. 3</u>
- 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]](https://tex.z-dn.net/?f=%5Cint%20%7Be%5E%7Bau%7Dsin%28bu%29%7D%20%5C%2C%20du%20%3D%20%5Cfrac%7B-e%5E%7Bau%7Dcos%28bu%29%7D%7Bb%7D%20%2B%20%5Cfrac%7Ba%7D%7Bb%7D%20%5B%5Cfrac%7Be%5E%7Bau%7Dsin%28bu%29%7D%7Bb%7D%20-%20%5Cfrac%7Ba%7D%7Bb%7D%20%5Cint%20%28%7Be%5E%7Bau%7D%20sin%28bu%29%7D%29%20%5C%2C%20du%5D)
- [Integral - Alg] Distribute Brackets:

- [Integral - Alg] Isolate Original Terms:

- [Integral - Alg] Rewrite:

- [Integral - Alg] Isolate Original:

- [Integral - Alg] Rewrite Fraction:

- [Integral - Alg] Combine Like Terms:

- [Integral - Alg] Divide:

- [Integral - Alg] Multiply:
![\int {e^{au}sin(bu)} \, du = \frac{1}{a^2+b^2} [ae^{au}sin(bu) - be^{au}cos(bu)]](https://tex.z-dn.net/?f=%5Cint%20%7Be%5E%7Bau%7Dsin%28bu%29%7D%20%5C%2C%20du%20%3D%20%5Cfrac%7B1%7D%7Ba%5E2%2Bb%5E2%7D%20%5Bae%5E%7Bau%7Dsin%28bu%29%20-%20be%5E%7Bau%7Dcos%28bu%29%5D)
- [Integral - Alg] Factor:
![\int {e^{au}sin(bu)} \, du = \frac{e^{au}}{a^2+b^2} [asin(bu) - bcos(bu)]](https://tex.z-dn.net/?f=%5Cint%20%7Be%5E%7Bau%7Dsin%28bu%29%7D%20%5C%2C%20du%20%3D%20%5Cfrac%7Be%5E%7Bau%7D%7D%7Ba%5E2%2Bb%5E2%7D%20%5Basin%28bu%29%20-%20bcos%28bu%29%5D)
- [Integral] Integration Constant:
![\int {e^{au}sin(bu)} \, du = \frac{e^{au}}{a^2+b^2} [asin(bu) - bcos(bu)] + C](https://tex.z-dn.net/?f=%5Cint%20%7Be%5E%7Bau%7Dsin%28bu%29%7D%20%5C%2C%20du%20%3D%20%5Cfrac%7Be%5E%7Bau%7D%7D%7Ba%5E2%2Bb%5E2%7D%20%5Basin%28bu%29%20-%20bcos%28bu%29%5D%20%2B%20C)
And we have proved the integration formula!
64+36=100 √100=10 (right ans=10)
1) surface of a rectangular prism=2(length x width)+2(length x height)+2(width x height)
Therefore:
148 cm²=2(5 cm x 4 cm)+2(5 cm x h)+2(4 cm x h)=
148 cm²=40 cm²+10 cm h+8 cm h
18 cm h=148 cm²-40 cm²
18 cm h=108 cm²
h=108 cm² / 18 cm=6 cm.
answer: height=6 cm
2)
Volume of a rectangular prism= length x width x height
therefore:
34 cm³=(1.7 cm)(0.5 cm) h
0.85 cm² h=34 cm³
h=34 cm³/0.85 cm²
h=40 cm.
answer: height=40 cm
3)
volume of a cylinder: πr²h
therefore.
118.79 ft³=πr²(5 ft)
r=√(118.79 ft³/5π ft)≈2.75 ft
answer: radius=2.75 ft
4)
Surface area of the pyramid with square base=4(A side)+A base
A side=(1/2)(8ft)(12 ft)=48 ft²
A base=(8 ft)(8 ft)=64 ft²
surface area=4(48 ft²)+64 ft²=256 ft²
Answer: the surface area of this pyramid would be 256 ft².
5)
surface of a cone=πrs+πr²
therefore:
radius=diameter/2=6.2 ft/2=3.1 ft
63.3 ft²=π(3.1 ft) s+π(3.1 ft)²
3.1π ft s=33.109 ft²
s=33.109 ft² /3.1π ft
s≈3.4 ft
Answer: the slant height would be 3.4 ft.
6)
volume of a square pyramid=(area of base x heigth)/3
therefore:
area of base=(6 ft)(6 ft)=36 ft²
126.97 ft³=36 ft² h /3
h=126.97 ft³/12 ft²=10.58 ft
answer: the height would be 10.58 ft.
7)
volume of a cone =(base x height)/3
base of a cone=πr²
therefore:
199.23 cm³=πr²(9 cm)/3
r=√(199.23 cm³ / 3π cm)≈4.6 cm
answer: the radius would be 4.6 cm.
Answer:
fge or fgc
I think that because its kinda like the same
A. He feels tired and sick because the focus on the mechanical measurements of stars is not satisfying to him.<span>
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