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2 years ago

What is the equation of this line in standard form

Mathematics

1 answer:

Yuri [45]2 years ago

3 0

Line joining (-3, -1) and (1/2, 2)

Point point form for a line is

(c-a)(y-b) = (d-b)(x-a)

(1/2 - - 3)(y - -1) = (2 - -1)(x - -3)

(7/2)(y+1)=3(x+3)

7(y+1)=6(x+3)

7 - 6(3) = 6x - 7y

6x - 7y = -11

Answer: second choice, 6x - 7y = -11

You might be interested in

Write an equation y is 20% what is x

lisabon 2012 [21]

Answer:

x + .20 = 1

Step-by-step explanation:

The equation would be x + .20= 1, or 1.00, because if you turn 20% into a decimal, it becomes .20. The total you are trying to find is 1 whole, in which this case it would simply be one as there is no other information. The x is first as if you had the .20 first, you would have to divide each side by .20, which would make x = 5, and that would not make sense.

Finding X: To find, subtract on each side by .20 due to the fact you are adding and need to do the opposite, .20-.20 cancels out and 1.00-.20 is .80, so you are left with x=.80

4 0

2 years ago

Please help quick 12 points write the ratio 7 to 8 as a fraction decimal and percent

OLEGan [10]

Answer: 7/8 0.875 87.5%

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

If -y-2x^3=Y^2 then find D^2y/dx^2 at the point (-1,-2) in simplest form

algol13

Answer:

$\frac{d^2y}{dx^2} = \frac{-4}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Factoring

Calculus

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

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

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

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

-y - 2x³ = y²

Rate of change of tangent line at point (-1, -2)

Step 2: Differentiate Pt. 1

Find 1st Derivative

Implicit Differentiation [Basic Power Rule]: $-y'-6x^2=2yy'$
[Algebra] Isolate y' terms: $-6x^2=2yy'+y'$
[Algebra] Factor y': $-6x^2=y'(2y+1)$
[Algebra] Isolate y': $\frac{-6x^2}{(2y+1)}=y'$
[Algebra] Rewrite: $y' = \frac{-6x^2}{(2y+1)}$

Step 3: Differentiate Pt. 2

Find 2nd Derivative

Differentiate [Quotient Rule/Basic Power Rule]: $y'' = \frac{-12x(2y+1)+6x^2(2y')}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x+12x^2y'}{(2y+1)^2}$
[Derivative] Back-Substitute y': $y'' = \frac{-24xy-12x+12x^2(\frac{-6x^2}{2y+1} )}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x-\frac{72x^4}{2y+1} }{(2y+1)^2}$

Step 4: Find Slope at Given Point

[Algebra] Substitute in x and y: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(-1)^4}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(1)}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Multiply: $y''(-1,-2) = \frac{-48+12-\frac{72}{-4+1} }{(-4+1)^2}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{(-3)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{9}$
[Pre-Algebra] Divide: $y''(-1,-2) = \frac{-36+24 }{9}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-12}{9}$
[Pre-Algebra] Simplify: $y''(-1,-2) = \frac{-4}{3}$

6 0

2 years ago

Give a verbal description of the following absolute value function f(x) = 3 | x - 5 | + 4

lisov135 [29]

Answer:

It is the distance of x from zero. Examples:The absolute value

of 5 is 5.

The absolute value of -10 is 10.

(Complex numbers) The absolute value of 4 + 3 is 5.

Step-by-step explanation:

7 0

3 years ago

What's the scientific notation of 0.000000109

Over [174]

0.000000109 in scientific notation will look as it follows:

1.09 * 10^-7

or

1.09 E-7

It's the same as 1.09 * (-10000000)

I hope it'll give you some clue of scientific notation with decimals.

5 0

3 years ago