PLEASE HELP! ILL MARK BRAINLIEST!!

HELLO!! I NEED HELP! Please show full Solutions. I will mark brainliest for the best answer. Thank you and god bless.

Greeley [361]

((x+2)(x+2))-((x-5)(1))=A

x^2+4x+4-x+5=A

x^2+3x+9=A

8 0

2 years ago

Which equations results from adding the equations in this system? x+6y=9 -x+2y=-15

UkoKoshka [18]

Answer:

8y = -6

Step-by-step explanation:

Simply add the equations as you would numbers.

The 'x's cancel each other out. The 'y's add together as normal numbers.

9 + (-15) = -6

Hope that this helps!

4 0

2 years ago

To solve 5/7 - 2/n = 1/21

Usimov [2.4K]

Answer:

the answer is n=3

Step-by-step explanation:

3 0

2 years ago

Which is a better deal 12.4 ounces for $4.99 or 21 ounces for $6.49?

ella [17]

Answer:

Getting 21 ounces for $6.49 is a better deal!

Step-by-step explanation:

4.99 / 12.4 = 0.402419...

6.49 / 21 = 0.30904...

0.4 is larger than 0.30 therefore 0.30 is less and a better deal.

Have an amazing day! ^-^

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%281-x%5E%7B2%7D%20%29%5E%7B3%2F2%7D%20%7D%20%5C%2C%20dx" id="TexFo

Ludmilka [50]

Answer: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}$ General Formulas and Concepts:

Pre-Calculus

Trigonometric Identities

Calculus

Differentiation

Derivatives
Derivative Notation

Integration

Integrals
Definite/Indefinite Integrals
Integration Constant C

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

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

U-Substitution

Trigonometric Substitution

Reduction Formula: $\displaystyle \int {cos^n(x)} \, dx = \frac{n - 1}{n}\int {cos^{n - 2}(x)} \, dx + \frac{cos^{n - 1}(x)sin(x)}{n}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution (trigonometric substitution).

Set u: $\displaystyle x = sin(u)$
[u] Differentiate [Trigonometric Differentiation]: $\displaystyle dx = cos(u) \ du$
Rewrite u: $\displaystyle u = arcsin(x)$

Step 3: Integrate Pt. 2

[Integral] Trigonometric Substitution: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[1 - sin^2(u)]^\Big{\frac{3}{2}} \, du$
[Integrand] Rewrite: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[cos^2(u)]^\Big{\frac{3}{2}} \, du$
[Integrand] Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos^4(u)} \, du$
[Integral] Reduction Formula: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{4 - 1}{4}\int \limits^a_b {cos^{4 - 2}(x)} \, dx + \frac{cos^{4 - 1}(u)sin(u)}{4} \bigg| \limits^a_b$
[Integral] Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4}\int\limits^a_b {cos^2(u)} \, du$
[Integral] Reduction Formula: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg|\limits^a_b + \frac{3}{4} \bigg[ \frac{2 - 1}{2}\int\limits^a_b {cos^{2 - 2}(u)} \, du + \frac{cos^{2 - 1}(u)sin(u)}{2} \bigg| \limits^a_b \bigg]$
[Integral] Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}\int\limits^a_b {} \, du + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg]$
[Integral] Reverse Power Rule: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}(u) \bigg| \limits^a_b + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg]$
Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3cos(u)sin(u)}{8} \bigg| \limits^a_b + \frac{3}{8}(u) \bigg| \limits^a_b$
Back-Substitute: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(arcsin(x))sin(arcsin(x))}{4} \bigg| \limits^a_b + \frac{3cos(arcsin(x))sin(arcsin(x))}{8} \bigg| \limits^a_b + \frac{3}{8}(arcsin(x)) \bigg| \limits^a_b$
Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x)}{8} \bigg| \limits^a_b + \frac{x(1 - x^2)^\Big{\frac{3}{2}}}{4} \bigg| \limits^a_b + \frac{3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b$
Rewrite: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x) + 2x(1 - x^2)^\Big{\frac{3}{2}} + 3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}$

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

Unit: Integration

Book: College Calculus 10e

8 0

2 years ago

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