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3 years ago

Which Relation is a function

Mathematics

1 answer:

ValentinkaMS [17]3 years ago

3 0

Answer:

It is A

Step-by-step explanation:

also i had the exact same question on a test which i got a 100% on.

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Which quadrant does the terminal side of angle -190° lie?

Luda [366]

Answer:

Quadrant lll

Step-by-step explanation:

Since the cosine of the angle is positive, the sine of the angle must be negative and the terminal side of the angle lies in either Quadrant III or IV.

8 0

3 years ago

Need help please its Calculus. Ill give the 5 stars as well.

algol13

Answer:

$\displaystyle y = 2e^\bigg{\frac{x^3}{3}} + 1$

General Formulas and Concepts:

Pre-Algebra

Order of Operations
Equality Properties

Algebra I

Functions
Function Notation
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Algebra II

Natural logarithms ln and Euler's number e

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Slope Fields

Separation of Variables

Solving Differentials

Integrals

Antiderivatives

Integration Constant C

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

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

U-Substitution

Logarithmic Integration: $\displaystyle \int {\frac{1}{u}} \, dx = ln|u| + C$

Step-by-step explanation:

*Note:

When solving differential equations in slope fields, disregard the integration constant C for variable y.

Step 1: Define

$\displaystyle \frac{dy}{dx} = x^2(y - 1)$

$\displaystyle f(0) = 3$

Step 2: Rewrite

Separation of Variables. Get differential equation to a form where we can integrate both sides and rewrite Leibniz Notation.

[Separation of Variables] Rewrite Leibniz Notation: $\displaystyle dy = x^2(y - 1) \ dx$
[Separation of Variables] Isolate y's together: $\displaystyle \frac{1}{y - 1} \ dy = x^2 \ dx$

Step 3: Find General Solution Pt. 1

[Differential] Integrate both sides: $\displaystyle \int {\frac{1}{y - 1}} \, dy = \int {x^2} \, dx$
[dx Integral] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle \int {\frac{1}{y - 1}} \, dy = \frac{x^3}{3} + C$

Step 4: Find General Solution Pt. 2

Identify variables for u-substitution for dy.

Set: $\displaystyle u = y - 1$
Differentiate [Basic Power Rule]: $\displaystyle du = dy$

Step 5: Find General Solution Pt. 3

[dy Integral] U-Substitution: $\displaystyle \int {\frac{1}{u}} \, du = \frac{x^3}{3} + C$
[dy Integral] Integrate [Logarithmic Integration]: $\displaystyle ln|u| = \frac{x^3}{3} + C$
[Equality Property] e both sides: $\displaystyle e^\bigg{ln|u|} = e^\bigg{\frac{x^3}{3} + C}$
Simplify: $\displaystyle |u| = Ce^\bigg{\frac{x^3}{3}}$
Rewrite: $\displaystyle u = \pm Ce^\bigg{\frac{x^3}{3}}$
Back-Substitute: $\displaystyle y - 1 = \pm Ce^\bigg{\frac{x^3}{3}}$
[Equality Property] Isolate y: $\displaystyle y = \pm Ce^\bigg{\frac{x^3}{3}} + 1$

General Form: $\displaystyle y = \pm Ce^\bigg{\frac{x^3}{3}} + 1$

Step 6: Find Particular Solution

Substitute in function values [General Form]: $\displaystyle 3 = \pm Ce^\bigg{\frac{0^3}{3}} + 1$
Simplify: $\displaystyle 3 = \pm C + 1$
[Equality Property] Isolate C: $\displaystyle 2 = \pm C$
Rewrite: $\displaystyle C = 2$
Substitute in C [General Form]: $\displaystyle y = 2e^\bigg{\frac{x^3}{3}} + 1$