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3 years ago

Rectangle ABCD has four right angles. 1. What is the sum of its angle measures?

Mathematics

2 answers:

sasho [114]3 years ago

3 0

90+90+90+90 is 360
Therefore, the interior angle sum of a rectangle is 360 degrees

Agata [3.3K]3 years ago

3 0

Answer:

Step-by-step explanation:

Step 1: 1 angle 90

4 angle -> 360

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I’m confused on this explain

romanna [79]

Triangles BAD and BCD are congruent by the HL theorem. That is, they have identical length hypotenuses (BD) and one leg is also identical in length to that in the other triangle. (AD ≅ CD)

Since the triangles are congruent, corresponding angles are equal, so ∠CBD = ∠ABD = 28°. The angle you are interested in is the sum of these two angles,

... ∠ABC = 28°+28° = 56°

6 0

3 years ago

-y- 3 (-3y+5) helpppppppp

valentinak56 [21]

Answer:

D = 8y-15

Step-by-step explanation:

-y-9y-15 than combine -y and -9y

so = 8y-15

3 0

3 years ago

Prove the following integration formula:

7nadin3 [17]

Answer:

See Explanation.

General Formulas and Concepts:

Pre-Algebra

Distributive Property
Equality Properties

Algebra I

Combining Like Terms
Factoring

Calculus

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:

Step 1: Define Integral

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

Step 2: Identify Variables Pt. 1

Using LIPET, we determine the variables for IBP.

Use Int Rules 2 + 3.

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

Step 3: Integrate Pt. 1

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

Step 4: Identify Variables Pt. 2

Using LIPET, we determine the variables for the 2nd IBP.

Use Int Rules 2 + 3.

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

Step 5: Integrate Pt. 2

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

Step 6: Integrate Pt. 3

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