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2 years ago

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The equation y + 6 = one-third (X minus 9) is written in point-slope form. What is the equation written in slope-intercept form?

y = one-third x minus 3 y = one-third x + 9 y = one-third x minus 9 y = one-third x + 3

1 answer:

stiks02 [169]2 years ago

4 0

The equation written in slope-intercept form is y = (1/3)x -9

<h3>What is a linear Equation ?</h3>

A linear equation is that which can be represented in the form of y =mx +c , m is the slope and c is the intercept.

The given equation is

y+6 = 1/3(x-9)

To determine the slope and the intercept this equation needs to be solved

y+6 = (1/3)x - 3

y = (1/3)x -9

Therefore this is the slope intercept format , here

the slope = 1/3

intercept = -9

To know more about Linear Equation

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Someone please help me :((

Dmitrij [34]

8x^2-3 = sqrt(16x+9)
[8x^2-3]^2 = [sqrt(16x+9)]^2 ... square both sides
64x^4-48x^2+9 = 16x+9
64x^4-48x^2+9-16x-9 = 0
64x^4-48x^2-16x = 0
16x(4x^3-3x-1) = 0
16x(x-1)(2x+1)^2 = 0
16x=0 or x-1=0 or (2x+1)^2 = 0
x=0 or x=1 or x = -1/2

The possible solutions are x=0 or x=1 or x = -1/2

We need to check all the possible solutions

Checking x=0
8x^2-3 = sqrt(16x+9)
8(0)^2-3 = sqrt(16*0+9)
-3 = 3
The equation is false so x=0 is extraneous (not a real solution)

Checking x=1
8x^2-3 = sqrt(16x+9)
8(1)^2-3 = sqrt(16*1+9)
5 = 5
Equation is true. The value x=1 is a solution

Checking x=-1/2
8x^2-3 = sqrt(16x+9)
8(-1/2)^2-3 = sqrt(16(-1/2)+9)
-1 = 1
The equation is false so x=-1/2 is extraneous (not a real solution)

Therefore, the only answer is choice A) 1

6 0

4 years ago

PLease answer !!! Find constants $A$ and $B$ such that \[\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\] for all

Xelga [282]

Answer:

1. $(A,B) = (3,-2)$

2. The values of t are: -3, -1

Step-by-step explanation:

Given

$\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}$

$|t| = 2t + 3$

Required

Solve for the unknown

Solving $\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}$

Take LCM

$\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{(x - 2)(x-1)}$

Expand the denominator

$\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{x^2 - 2x + x -2}$

$\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{x^2 - x -2}$

Both denominators are equal; So, they can cancel out

$x + 7 = A(x+1) + B(x-2)$

Expand the expression on the right hand side

$x + 7 = Ax + A + Bx - 2B$

Collect and Group Like Terms

$x + 7 = (Ax + Bx) + (A - 2B)$

$x + 7 = (A + B)x + (A - 2B)$

By Direct comparison of the left hand side with the right hand side

$(A + B)x = x$

$A - 2B = 7$

Divide both sides by x in $(A + B)x = x$

$A + B = 1$

Make A the subject of formula

$A = 1 - B$

Substitute 1 - B for A in $A - 2B = 7$

$1 - B - 2B = 7$

$1 - 3B = 7$

Subtract 1 from both sides

$1 - 1 - 3B = 7 - 1$

$-3B = 6$

Divide both sides by -3

$B = -2$

Substitute -2 for B in $A = 1 - B$

$A = 1 - (-2)$

$A = 1 + 2$

$A = 3$

Hence;

$(A,B) = (3,-2)$

Solving $|t| = 2t + 3$

Because we're dealing with an absolute function; the possible expressions that can be derived from the above expression are;

$t = 2t + 3$ and $-t = 2t + 3$

Solving $t = 2t + 3$

Make t the subject of formula

$t - 2t = 3$

$-t = 3$

Multiply both sides by -1

$t = -3$

Solving $-t = 2t + 3$

Make t the subject of formula

$-t - 2t = 3$

$-3t = 3$

Divide both sides by -3

$t = -1$

<em>Hence, the values of t are: -3, -1</em>

7 0

3 years ago

Read 2 more answers

Help please I got to pass

mixer [17]

Answer:

A'(5,3)

Step-by-step explanation:

First, you must understand that A(5,3) is the pre-image and that A' is what we are looking for which is the image.

With that in mind, you translate the preimage by adding or subtracting from the x and y values.

(x+4, y-3)

To find the x value of the pre-image, you will add 4 to the preimages' x value

Pre-image: A(1,6)

To find the y value, you subtract the preimages' y value by 3.

Hope this helps!

4 0

3 years ago

Help on this please, i dont get this problem.

Ahat [919]

Answer:

<h2>A = 37 cm²</h2>

Step-by-step explanation:

Look at the picture.

We have two rectangles:

15cm × 2cm and 7cm × 1cm.

Calculate the areas:

A₁ = 15 · 2 = 30 cm²

A₂ = 7 · 1 = 7 cm²

The area of the figure:

A = A₁ + A₂ → A = 30 cm² + 7 cm² = 37 cm²

5 0

3 years ago

(8.11×107)–(4.3×107)

stira [4]

Put it in a calculator bro

4 0

3 years ago

Read 2 more answers