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3 years ago

Help I will mark Brainly

Mathematics

1 answer:

Dmitry_Shevchenko [17]3 years ago

4 0

I believe that the column with gluten free you would equally divide the amount of people so each row would have 6 and in the column for not gluten free you would divide the total by 2 and get 58 for both positive and negative in that column

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Doug read 231 pages in 7 hours.he read the same number of pages each hour.how many pages did he read in 1 hour

tino4ka555 [31]

If we let $p$ be the number of pages read in 1 hour, then in 7 hours Doug reads $7p$ pages. We know that this number of pages is 231, so we have

$7p=231 \iff p = \dfrac{231}{7} = 33$

8 0

4 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

The ellipse 14x2+2x+y2=114x2+2x+y2=1 has its center at the point (b,c)(b,c) where

Nana76 [90]

18x^2 + 2x + y^2 = 1
==> 18(x^2 + x/9) + y^2 = 1 
==> 18(x^2 + x/9 + 1/324) + y^2 = 1 + 1/18 
==> 18(x + 1/18)^2 + y^2 = 19/18 
==> (x + 1/18)^2/19 + (y - 0)^2/(19/18) = 1. 

By comparing this to the standard form of an ellipse, the center is at (-1/18, 0).

6 0

3 years ago

PLZ HELP ITS FOR MY EXAM !! What information can the discriminant tell you about a quadratic relation in the form y = ax² + bx +

Romashka [77]

This equals a symmetrical

7 0

3 years ago

On a piece of paper, graph y≤ -2x.Then determine which answer matches the graph you drew.

Solnce55 [7]

D. (-2,-6) (1,3) is the correct answer I believe

5 0

4 years ago

Read 2 more answers