All categories

2 years ago

PLEASE HELP, MARKING BRAINLIEST!! if 4m = 9n, then m/n=

Mathematics

1 answer:

kondaur [170]2 years ago

7 0

Answer:

m/n=9/4

Step-by-step explanation:

4m = 9n

m/n=9/4

You might be interested in

Segment BDSegment BD is a perpendicular bisector of segment ACsegment AC . AD=8x−15AD=8x−15 and DC=57DC=57 .

Charra [1.4K]

I hope this helps you

7 0

3 years ago

Read 2 more answers

What is another name for a relation that has each element in its domain paired with exactly one element in its range? Apex

8090 [49]

Another name for a relation is a function.

7 0

3 years ago

I need help please.!

gulaghasi [49]

Hey ! I know for sure A is one of the correct answers i’m working on the rest now

8 0

3 years ago

Fill in the blank to make the fractions equivalent 2/3 = Blank/15

kotegsom [21]

Answer: 2/3 = 10/15

8 0

2 years ago

Find the area of the region enclosed by the graphs of the functions

Vaselesa [24]

Answer:

$\displaystyle A = \frac{8}{21}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Terms/Coefficients
Functions
Function Notation
Graphing
Solving systems of equations

Calculus

Area - Integrals

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

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

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

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

*Note:

Remember that for the Area of a Region, it is top function minus bottom function.

Step 1: Define

f(x) = x²

g(x) = x⁶

Bounded (Partitioned) by x-axis

Step 2: Identify Bounds of Integration

Find where the functions intersect (x-values) to determine the bounds of integration.

Simply graph the functions to see where the functions intersect (See Graph Attachment).

Interval: [-1, 1]

Lower bound: -1

Upper Bound: 1

Step 3: Find Area of Region

Integration

Substitute in variables [Area of a Region Formula]: $\displaystyle A = \int\limits^1_{-1} {[x^2 - x^6]} \, dx$
[Area] Rewrite [Integration Property - Subtraction]: $\displaystyle A = \int\limits^1_{-1} {x^2} \, dx - \int\limits^1_{-1} {x^6} \, dx$
[Area] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle A = \frac{x^3}{3} \bigg| \limit^1_{-1} - \frac{x^7}{7} \bigg| \limit^1_{-1}$
[Area] Evaluate [Integration Rule - FTC 1]: $\displaystyle A = \frac{2}{3} - \frac{2}{7}$
[Area] Subtract: $\displaystyle A = \frac{8}{21}$

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

Unit: Area Under the Curve - Area of a Region (Integration)

Book: College Calculus 10e

6 0

3 years ago