Two thirds a number x plus 3 is 6

A transformation T : (x, y) → (x + 3, y + 1). Find the preimage of the point (4, 3) under the given transformation. (7, 4) (1, 2

motikmotik

Answer:

(1, 2)

Step-by-step explanation:

Remember that the final shape and position of a figure after a transformation is called the image, and the original shape and position of the figure is the pre-image.

In our case, our figure is just a point. We know that after the transformation T : (x, y) → (x + 3, y + 1), our image has coordinates (4, 3).

The transformation rule T : (x, y) → (x + 3, y + 1) means that we add 3 to the x-coordinate and add 1 to the y-coordinate of our pre-image. Now to find the pre-image of our point, we just need to reverse those operations; in other words, we will subtract 3 from the x-coordinate and subtract 1 from the y-coordinate.

So, our rule to find the pre-image of the point (4, 3) is:

T : (x, y) → (x - 3, y - 1)

We know that the x-coordinate of our image is 4 and its y-coordinate is 3.

Replacing values:

(4 - 3, 3 - 1)

(1, 2)

We can conclude that our pre-image is the point (1, 2).

6 0

3 years ago

*Brainly* Find the equation of the line shown in the graph. Write the equation in slope-intercept form.

Brums [2.3K]

Answer:

y = (-4/3)x + 6

Step-by-step explanation:

The Y-intercept of the graph is 6. You can find the slope by following rise over run, or the change in y divided by the change in x. From (0,6) to (3,2), there is a negative 4 change in y and positive 3 change in x.

3 0

3 years ago

The solutions to (x + 3)^2- 4=0 are x = -1 and x = -5<br><br> True or false

Neko [114]

Answer:

THE ANSWER WOULD BE TRUE MY FRIEND

6 0

2 years ago

A product can be made in sizes huge, average and tiny which yield a net unit profit of $14, $10, and$5, respectively. Three cent

navik [9.2K]

Answer:

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃ to maximize

Subject to:

First center X₁₁ + X₂₁ + X₃₁ ≤ 550

Second center X₁₂ + X₂₂ + X₃₂ ≤ 750

Third center X₁₃ + X₂₃ + X₃₃ ≤ 275

22* X₁₁ + 16* X₂₁ + 9*X₃₁ ≤ 11000

22* X₁₂ + 16* X₂₂ + 9*X₃₂ ≤ 2700

22*X₁₃ + 16* X₂₃ + 9*X₃₃ ≤ 3400

X₁₁ + X₁₂ + X₁₃ ≤ 710

X₂₁ + X₂₂ + X₂₃ ≤ 900

X₃₁ + X₃₂ + X₃₃ ≤ 350

2700*(X₁₁ + X ₂₁ + X₃₁) - 11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁ + X ₂₁ + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

Step-by-step explanation:

Let´s call Xij product size i produced in center j

According to this, we get the following set of variable

X₁₁ product size huge produced in center 1

X₁₂ product size huge produced in center 2

X₁₃ product size huge produced in center 3

X₂₁ product size average produced in center 1

X₂₂ product size average produced in center 2

X₂₃ product size average produced in center 3

X₃₁ product size-tiny produced in center 1

X₃₂ product size-tiny produced in center 2

X₃₃ product size-tiny produced in center 3

Then Objective function is

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Constrains

Center capacity

1.- First center X₁₁ + X₂₁ + X₃₁ ≤ 550

2.- Second center X₁₂ + X₂₂ + X₃₂ ≤ 750

3.- Third center X₁₃ + X₂₃ + X₃₃ ≤ 275

Water available

1.- 22* X₁₁ + 16* X₂₁ + 9*X₃₁ ≤ 11000

2.- 22* X₁₂ + 16* X₂₂ + 9*X₃₂ ≤ 2700

3.- 22*X₁₃ + 16* X₂₃ + 9*X₃₃ ≤ 3400

Demand constrain

Product huge

X₁₁ + X₁₂ + X₁₃ ≤ 710

Product average

X₂₁ + X₂₂ + X₂₃ ≤ 900

Product tiny

X₃₁ + X₃₂ + X₃₃ ≤ 350

Fraction SP/CC must be the same

First and second centers fraction SP/CC

(X₁₁ + X ₂₁ + X₃₁)/ 11000 = (X₁₂ + X₂₂ + X₃₂)/ 2700

2700*(X₁₁ + X ₂₁ + X₃₁) - 11000*(X₁₂ + X₂₂ + X₃₂) = 0

First and third centers fraction SP/CC

(X₁₁ + X ₂₁ + X₃₁)/ 11000 = ( X₁₃ + X₂₃ + X₃₃)/ 3400

3400*(X₁₁ + X ₂₁ + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Subject to:

First center X₁₁ + X₂₁ + X₃₁ ≤ 550

Second center X₁₂ + X₂₂ + X₃₂ ≤ 750

Third center X₁₃ + X₂₃ + X₃₃ ≤ 275

22* X₁₁ + 16* X₂₁ + 9*X₃₁ ≤ 11000

22* X₁₂ + 16* X₂₂ + 9*X₃₂ ≤ 2700

22*X₁₃ + 16* X₂₃ + 9*X₃₃ ≤ 3400

X₁₁ + X₁₂ + X₁₃ ≤ 710

X₂₁ + X₂₂ + X₂₃ ≤ 900

X₃₁ + X₃₂ + X₃₃ ≤ 350

2700*(X₁₁ + X ₂₁ + X₃₁) - 11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁ + X ₂₁ + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

6 0

3 years ago

After being filled, a basketball loses 3.2% of its air every day. Considering that the initial volume of the ball was 840 cubic

Sliva [168]

Answer:

volume = 840 cubic cm

(3.2/100)×840= 26.88 per day

after 4 days = 26.88×4=107.5

answer is 107.5

8 0

1 year ago

Read 2 more answers