50 POINTS !! PLEASE HELP !! ILL GIVE BRAINLIEST TO THE RIGHT ANSWERS.

Which ordered pair describes the location of point A? (-6, 5) (-6, -5) (5, -6) (5, 6)

Neko [114]

Answer:

the answer is -6,5

Step-by-step explanation:

just look at the left side ( the x axis )first then right(the y axis)

8 0

2 years ago

The area of a rectangle is 162/3 ft2 . The length is 61/4 . Find the width. Show all work

brilliants [131]

Answer:

I have gotten 216/61, but I am not sure if this is right.

Step-by-step explanation:

By using the "Keep It, Change It, Flip It" tactic, I was able to come up with an answer and simplify it.

1. Keep the original divisor.

162/3

2. Change the sign.

Division becomes multiplication.

3. The last fraction is reciprocated.

61/4 becomes 4/61.

3 0

2 years ago

I need answer ASAP plz!!!

ANTONII [103]

It’s B I hope this helps :)

6 0

2 years ago

Show whether or not y=x+3 is tangential to the curve y^2=x

wolverine [178]

The line y = x + 3 has slope 1, so we look for points on the curve where the tangent line, whose slope is dy/dx, is equal to 1.

y² = x

Take the derivative of both sides with respect to x, assuming y = y(x) :

2y dy/dx = 1

dy/dx = 1/(2y)

Solve for y when dy/dx = 1 :

1 = 1/(2y)

2y = 1

y = 1/2

When y = 1/2, we have x = y² = (1/2)² = 1/4. However, for the given line, when y = 1/2, we have x = y - 3 = 1/2 - 3 = -5/2.

This means the line y = x + 3 is not a tangent to the curve y² = x. In fact, the line never even touches y² = x :

x = y² ⇒ y = y² + 3 ⇒ y² - y + 3 = 0

has no real solution for y.

3 0

1 year 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

2 years ago