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3 years ago

Write the linear equation in slope-intercept form 2x - 3y = 6

1 answer:

7 0

2x - 3y = 6
3y = -2x + 6

y = - $\frac{2}{3}$ x + 2

m = - $\frac{2}{3}$
c = 2

Linear Equation:
y = - $\frac{2}{3}$ x + 2

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Please please help me

Katarina [22]

Answer:

y = - 2x + 6

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (- 8, 22) and (x₂, y₂ ) = (3, 0)

m = $\frac{0-22}{3+8}$ = $\frac{-22}{11}$ = - 2, thus

y = - 2x + c ← is the partial equation

To find c substitute either of the 2 points into the partial equation

Using (3, 0), then

0 = - 6 + c ⇒ c = 0 + 6 = 6

y = - 2x + 6 ← equation of line

5 0

3 years ago

-2(x + 5) = -16 = ???

34kurt

Answer:

if you're looking for x its -13

8 0

3 years ago

Read 2 more answers

Need answers fast please answer

valina [46]

I would say (-1,-1) because it’s on the dotted line and all the other points are within the orange highlight

7 0

2 years ago

Solve for x.

Ivahew [28]

ANSWER

$x=\frac{2-\sqrt{10}} {3}$

or

$x=\frac{\sqrt{10}+2} {3}$

We have

$3x^2-4x-2=0$

Since we cannot factor easily, we complete the square.

Adding 2 to both sides give,

$3x^2-4x=2$

Dividing through by 3 gives

$x^2-\frac{4}{3}x= \frac{2}{3}$

Adding $(-\frac{2}{3})^2$ to both sides gives

$x^2-\frac{4}{3}x+(-\frac{2}{3})^2= \frac{2}{3}+(-\frac{2}{3})^2$

The expression on the Left Hand side is a perfect square.

$(x-\frac{2}{3})^2= \frac{2}{3}+\frac{4}{9}$

$\Rightarrow (x-\frac{2}{3})^2= \frac{10}{9}$

$\Rightarrow (x-\frac{2}{3})=\pm \sqrt{\frac{10}{9}}$

$\Rightarrow (x)=\frac{2}{3} \pm {\frac{\sqrt{10}}{3}$

Splitting the plus or minus sign gives

$x=\frac{2- \sqrt{10}} {3}$

or

$x=\frac{\sqrt{10}+2} {3}$

3 0

3 years ago

Read 2 more answers

Solve the Anti derivative.

Alex Ar [27]

Answer:

$\displaystyle \int {\frac{1}{9x^2+4}} \, dx = \frac{1}{6}arctan(\frac{3x}{2}) + C$

General Formulas and Concepts:

Algebra I

Factoring

Calculus

Antiderivatives - integrals/Integration

Integration Constant C

U-Substitution

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Trig Integration: $\displaystyle \int {\frac{du}{a^2 + u^2}} = \frac{1}{a}arctan(\frac{u}{a}) + C$

Step-by-step explanation:

Step 1: Define

$\displaystyle \int {\frac{1}{9x^2 + 4}} \, dx$

Step 2: Integrate Pt. 1

[Integral] Factor fraction denominator: $\displaystyle \int {\frac{1}{9(x^2 + \frac{4}{9})}} \, dx$
[Integral] Integration Property - Multiplied Constant: $\displaystyle \frac{1}{9} \int {\frac{1}{x^2 + \frac{4}{9}}} \, dx$

Step 3: Identify Variables

Set up u-substitution for the arctan trig integration.

$\displaystyle u = x \\ a = \frac{2}{3} \\ du = dx$

Step 4: Integrate Pt. 2

[Integral] Substitute u-du: $\displaystyle \frac{1}{9} \int {\frac{1}{u^2 + (\frac{2}{3})^2} \, du$
[Integral] Trig Integration: $\displaystyle \frac{1}{9}[\frac{1}{\frac{2}{3}}arctan(\frac{u}{\frac{2}{3}})] + C$
[Integral] Simplify: $\displaystyle \frac{1}{9}[\frac{3}{2}arctan(\frac{3u}{2})] + C$
[integral] Multiply: $\displaystyle \frac{1}{6}arctan(\frac{3u}{2}) + C$
[Integral] Back-Substitute: $\displaystyle \frac{1}{6}arctan(\frac{3x}{2}) + C$

Topic: AP Calculus AB

Unit: Integrals - Arctrig

Book: College Calculus 10e

7 0

2 years ago