A turtle pack that normally sells for $39 is on sale for 33% off. Find the

By use of technology, we investigated Mary’s investment and created the model, M(x) = 3.03(1.28)2x, in thousands of dollars. Wha

aleksandr82 [10.1K]

The initial investment was 3.03 because this is the value that does not change and comes before the value with the exponent "2x" is mentioned.

6 0

3 years ago

Read 2 more answers

Sean bought a 6 inch sub. He ate 3 3/4 inches. How much sub does he have left in the simplest form.

irinina [24]

Answer: 2 1/2

hope this helps

4 0

3 years ago

The equation of a circle is given below.

BARSIC [14]

Given:

The equation of the circle is $x^2+(y+4)^2=64$

We need to determine the center and radius of the circle.

Center:

The general form of the equation of the circle is $(x-h)^2+(y-k)^2=r^2$

where (h,k) is the center of the circle and r is the radius.

Let us compare the general form of the equation of the circle with the given equation $x^2+(y+4)^2=64$ to determine the center.

The given equation can be written as,

$(x-0)^2+(y+4)^2=64$

Comparing the two equations, we get;

(h,k) = (0,-4)

Therefore, the center of the circle is (0,-4)

Radius:

Let us compare the general form of the equation of the circle with the given equation $x^2+(y+4)^2=64$ to determine the radius.

Hence, the given equation can be written as,

$x^2+(y+4)^2=8^2$

Comparing the two equation, we get;

$r^2=8^2$

$r=8$

Thus, the radius of the circle is 8

3 0

3 years ago

Cual es el resultado de simplificar. Adjunto foto

Margaret [11]

Answer:

⣿⣯⣿⣟⣟⡼⣿⡼⡿⣷⣿⣿⣿⠽⡟⢋⣿⣿⠘⣼⣷⡟⠻⡿⣷⡼⣝⡿⡾⣿ ⣿⣿⣿⣿⢁⣵⡇⡟⠀⣿⣿⣿⠇⠀⡇⣴⣿⣿⣧⣿⣿⡇⠀⢣⣿⣷⣀⡏⢻⣿ ⣿⣿⠿⣿⣿⣿⠷⠁⠀⠛⠛⠋⠀⠂⠹⠿⠿⠿⠿⠿⠉⠁⠀⠘⠛⠛⠛⠃⢸⣯ ⣿⡇⠀⣄⣀⣀⣈⣁⠈⠉⠃⠀⠀⠀⠀⠀⠀⠀⠀⠠⠎⠈⠀⣀⣁⣀⣀⡠⠈⠉ ⣿⣯⣽⡿⢟⡿⠿⠛⠛⠿⣶⣄⠀⠀⠀⠀⠀⠀⠈⢠⣴⣾⠛⠛⠿⠻⠛⠿⣷⣶ ⣿⣿⣿⠀⠀⠀⣿⡿⣶⣿⣫⠉⠀⠀⠀⠀⠀⠀⠀⠈⠰⣿⠿⠾⣿⡇⠀⠀⢺⣿ ⣿⣿⠻⡀⠀⠀⠙⠏⠒⡻⠃⠀⠀⠀⠀⣀⠀⠀⠀⠀⠀⠐⡓⢚⠟⠁⠀⠀⡾⢫ ⣿⣿⠀⠀⡀⠀⠀⡈⣉⡀⡠⣐⣅⣽⣺⣿⣯⡡⣴⣴⣔⣠⣀⣀⡀⢀⡀⡀⠀⣸ ⣿⣿⣷⣿⣟⣿⡿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⢻⢾⣷⣿ ⣿⣿⣟⠫⡾⠟⠫⢾⠯⡻⢟⡽⢶⢿⣿⣿⡛⠕⠎⠻⠝⠪⢖⠝⠟⢫⠾⠜⢿⣿ ⣿⣿⣿⠉⠀⠀⠀⠀⠈⠀⠀⠀⠀⣰⣋⣀⣈⣢⠀⠀⠀⠀⠀⠀⠀⠀⠀⣐⢸⣿ ⣿⣿⣿⣆⠀⠀⠀⠀⠀⠀⠀⠀⢰⣿⣿⣿⣿⣿⣧⠀⠀⠀⠀⠀⠀⠀⠀⢀⣾⣿ ⣿⣿⣿⣿⣦⡔⠀⠀⠀⠀⠀⠀⢻⣿⡿⣿⣿⢽⣿⠀⠀⠀⠀⠀⠀⠀⣠⣾⣿⣿ ⣿⣿⣿⣿⣿⣿⣶⣤⣀⠀⠀⠀⠘⠛⢅⣙⣙⠿⠉⠀⠀⠀⢀⣠⣴⣿⣿⣿⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣶⣤⣄⣅⠀⠓⠀⠀⣀⣠⣴⣺⣿⣿⣿⣿⣿⣿⣿⣿

Step-by-step explanation:

i nu speak that language

6 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

3 years ago

Read 2 more answers