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3 years ago

Can someone help me

Mathematics

1 answer:

IceJOKER [234]3 years ago

3 0

Answer:

Slope=m=5/4

Point through which line is passing is (X,Y)=(4,−1)

By using slope-point form, the equation of line is given as,

(y−Y)=m(x−X)

(y−(−1))=5/4(x−4)

(y+1)=5x/4−5

(y+1)=(5x−20)/4

4(y+1)=5x−20

4y+4=5x−20

5x−4y−20−4=0

5x−4y−24=0

is the equation of line passing through the point (4,−1)and having slope 5/4

Hope it helps you :)

Step-by-step explanation:

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3 years ago

The length of a rectangle is increasing at a rate of 4 meters per day and the width is increasing at a rate of 1 meter per day.

puteri [66]

Answer:

$\displaystyle \frac{dA}{dt} = 102 \ m^2/day$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Geometry

Area of a Rectangle: A = lw

l is length
w is width

Calculus

Derivatives

Derivative Notation

Implicit Differentiation

Differentiation with respect to time

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

Step-by-step explanation:

Step 1: Define

$\displaystyle l = 10 \ meters$

$\displaystyle \frac{dl}{dt} = 4 \ m/day$

$\displaystyle w = 23 \ meters$

$\displaystyle \frac{dw}{dt} = 1 \ m/day$

Step 2: Differentiate

[Area of Rectangle] Product Rule: $\displaystyle \frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$

Step 3: Solve

[Rate] Substitute in variables [Derivative]: $\displaystyle \frac{dA}{dt} = (10 \ m)(1 \ m/day) + (23 \ m)(4 \ m/day)$
[Rate] Multiply: $\displaystyle \frac{dA}{dt} = 10 \ m^2/day + 92 \ m^2/day$
[Rate] Add: $\displaystyle \frac{dA}{dt} = 102 \ m^2/day$