PLEASE HELP ME!

three brothers and a sister shared a sum of money equally among themselves. if the brothers alone had shared the money, then the

Virty [35]

Answer:

1. t=240

2.t/3=(t/4)+20

Step-by-step explanation:

4 0

2 years ago

To which graph does the point (−1, 3) belong? y ≥ 2x + 6 y ≥ 3x − 1 y ≥ −x + 3 y ≥ −2x + 5

Anna35 [415]

Well first find the slope of the point
y=mx+b y=-1x+3
slope = -1
then put the problem in point slope form y - y1 = m (x-x1)
y + 3 = -1 (x+1) then solve for y=mx+b again
so y> 2x + 6 is the answer

4 0

2 years ago

Read 2 more answers

Hello again! This is another Calculus question to be explained.

podryga [215]

Answer:

See explanation.

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions

Function Notation

Exponential Property [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the following and are trying to find the second derivative at x = 2:

$\displaystyle f(2) = 2$

$\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}$

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

$\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}$

When we differentiate this, we must follow the Chain Rule: $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]$

Use the Basic Power Rule:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')$

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]$

Simplifying it, we have:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]$

We can rewrite the 2nd derivative using exponential rules:

$\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}$

To evaluate the 2nd derivative at x = 2, simply substitute in x = 2 and the value f(2) = 2 into it:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}$

When we evaluate this using order of operations, we should obtain our answer:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219$

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

Unit: Differentiation

5 0

2 years ago

A robot is expected to filter pollution out of at least 350 liters of air and water. It filters air at the rate of 50 liters per

s2008m [1.1K]

Answer:

$20W+50A\geq 350$

Step-by-step explanation:

we are given that A robot is expected to filter pollution out of at least 350 liters of air and water.

Also It filters air at the rate of 50 liters per minute, and it filters water at the rate of 20 liters per minute.

The inequality for number of minutes the robot should filter air (A) and water (W) to meet this expectations can be writte as follows:

$20W+50A\geq 350$

Hence the required inequality has been formulated.

4 0

2 years ago

Read 2 more answers

Pls help (have a good day Jesus loves you:) :) .)

murzikaleks [220]

Answer:

4^6

Step-by-step explanation:

5 0

2 years ago