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3 years ago

What is the slope of the line that is perpendicular to the line whose equation is 2x+y=4??

Mathematics

1 answer:

zmey [24]3 years ago

6 0

Answer:

The slope would be 1/2

Step-by-step explanation:

y= mx +b

m= slope

perpendicular lines have slopes that are multiplied in order to get -1

these are important

2x + y = 4

subtract 2x

y=-2x +4

your slope is -2

-2 * ? = -1

? = -1/-2

? = 1/2

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Enter an expression equivalent to (3x^2+2y^2-3x) + (2x^2+y^2-2x) - (x^2 + 3y^2 +x)

11111nata11111 [884]

Answer:

4x² - 6x

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Distributive Property

Algebra I

Terms/Coefficients

Step-by-step explanation:

Step 1: Define

(3x² + 2y² - 3x) + (2x² + y² - 2x) - (x² + 3y² + x)

Step 2: Simplify

[Distributive Property] Distribute negative: 3x² + 2y² - 3x + 2x² + y² - 2x - x² - 3y² - x
Combine like terms (x²): 4x² + 2y² - 3x + y² - 2x - 3y² - x
Combine like terms (y²): 4x² - 3x - 2x - x
Combine like terms (x): 4x² - 6x

4 0

3 years ago

The radius of a cone is decreasing at a constant rate of 7 inches per second, and the volume is decreasing at a rate of 948 cubi

inessss [21]

Answer:

The height of cone is decreasing at a rate of 0.085131 inch per second.

Step-by-step explanation:

We are given the following information in the question:

The radius of a cone is decreasing at a constant rate.

$\displaystyle\frac{dr}{dt} = -7\text{ inch per second}$

The volume is decreasing at a constant rate.

$\displaystyle\frac{dV}{dt} = -948\text{ cubic inch per second}$

Instant radius = 99 inch

Instant Volume = 525 cubic inches

We have to find the rate of change of height with respect to time.

Volume of cone =

$V = \displaystyle\frac{1}{3}\pi r^2 h$

Instant volume =

$525 = \displaystyle\frac{1}{3}\pi r^2h = \frac{1}{3}\pi (99)^2h\\\\\text{Instant heigth} = h = \frac{525\times 3}{\pi(99)^2}$

Differentiating with respect to t,

$\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)$

Putting all the values, we get,

$\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)\\\\-948 = \frac{1}{3}\pi\bigg(2(99)(-7)(\frac{525\times 3}{\pi(99)^2}) + (99)(99)\frac{dh}{dt}\bigg)\\\\\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi} = (99)^2\frac{dh}{dt}\\\\\frac{1}{(99)^2}\bigg(\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi}\bigg) = \frac{dh}{dt}\\\\\frac{dh}{dt} = -0.085131$