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oksano4ka [1.4K]
3 years ago
9

What is the value of y if (4y + 8) - (7y - 12) = 11? (PLEASE ANSWER + 10 POINTS!!!!)?

Mathematics
1 answer:
lyudmila [28]3 years ago
6 0
Hello!

(4y + 8) - (7y - 12) = 11     is     1(4y + 8) - 1(7y - 12) = 11

1(4y + 8) - 1(7y - 12) = 11               Given
        4y + 8 - 7y + 12 = 11              Distribute the 1 and the -1
                    -3y + 20 = 11             Combine like terms
                            -3y = -9              Subtract 20 from both sides
                               y = 3                Divide both sides by -3

Answer:
y = 3
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the length of a sandbox is three feet longer than it’s width. Write the expression that would represent the area of the sandbox.
Rashid [163]

Part A : D.)

Part B : Length of the sandbox is 10 feet.

Step-by-step explanation:

Given,

Perimeter = 29 ft

We need to find the equation for the perimeter and also the length of the sandbox.

Solution,

Let the width of the sandbox be 'w'.

Now as per question said;

The length of the sandbox is 1 foot longer than twice the width of the sandbox.

So we can say that;

Length =

Now we know that the perimeter is equal to the sum of twice of length and width.

framing in equation form, we get;

Perimeter =

we have given the perimeter, so on substituting the value, we get;

Hence The equation used to find the width is  .

Now we solve for 'w'.

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Dividing both side by 6 we get;

Width of the sandbox = 4.5 ft

Length of the sandbox =

Hence Length of the sandbox is 10 feet.

6 0
3 years ago
Given: 2x^2+3x-5, g(x)=x+9. Find: (fg)(3)
Svetradugi [14.3K]

Answer:

264

Step-by-step explanation:

f(x)=2x^2+3x-5

g(x)=x+9

(fg)(x)=f(x) \times g(x)

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\implies (fg)(3)=22 \times 12

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8 0
2 years ago
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SashulF [63]

Answer:

7.211103

Step-by-step explanation:

For:

(X1, Y1) = (-3, 4)

(X2, Y2) = (1, -2)

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3 years ago
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Answer:

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3 years ago
Help me on this please
zalisa [80]

Answer:

1. (x, y) → (x + 3, y - 2)

Vertices of the image

a) (-2, - 3)

b) (-2, 3)

c) (2, 2)

2. (x, y) → (x - 3, y + 5)

Vertices of the image

a) (-3, 2)

b) (0, 2)

c) (0, 4)

d) (2, 4)

3. (x, y) → (x + 4, y)

Vertices of the image

a) (-1, -2)

b) (1, -2)

c) (3, -2)

4. (x, y) → (x + 6, y + 1)

Vertices of the image

a) (1, -1)

b) (1, -2)

c) (2, -2)

d) (2, -4)

e) (3, -1)

f) (3, -3)

g) (4, -3)

h) (1, -4)

5. (x, y) → (x, y - 4)

Vertices of the image

a) (0, -2)

b) (0, -3)

c) (2, -2)

d) (2, -4)

6. (x, y) → (x - 1, y + 4)

Vertices of the image

a) (-5, 3)

b) (-5, -1)

c) (-3, 0)

d) (-3, -1)

Explanation:

To identify each <u><em>IMAGE</em></u> you should perform the following steps:

  • List the vertex points of the preimage (the original figure) as ordered pairs.
  • Apply the transformation rule to every point of the preimage
  • List the image of each vertex after applying each transformation, also as ordered pairs.

<u>1. (x, y) → (x + 3, y - 2)</u>

The rule means that every point of the preimage is translated three units to the right and 2 units down.

Vertices of the preimage      Vertices of the image

a) (-5,2)                                   (-5 + 3, -1 - 2) = (-2, - 3)

b) (-5, 5)                                  (-5 + 3, 5 - 2) = (-2, 3)

c) (-1, 4)                                   (-1 + 3, 4 - 2) = (2, 2)

<u>2. (x,y) → (x - 3, y + 5)</u>

The rule means that every point of the preimage is translated three units to the left and five units down.

Vertices of the preimage      Vertices of the image

a) (0, -3)                                   (0 - 3, -3 + 5) = (-3, 2)

b) (3, -3)                                   (3 - 3, -3  + 5) = (0, 2)

c) (3, -1)                                    (3 - 3, -1 + 5) = (0, 4)

d) (5, -1)                                    (5 - 3, -1 + 5) = (2, 4)

<u>3. (x, y) → (x + 4, y)</u>

The rule represents a translation 4 units to the right.

Vertices of the preimage   Vertices of the image

a) (-5, -2)                               (-5 + 4, -2) = (-1, -2)

b) (-3, -5)                               (-3 + 4, -2) = (1, -2)

c) (-1, -2)                                (-1 + 4, -2) = (3, -2)

<u>4. (x, y) → (x + 6, y + 1)</u>

Vertices of the preimage      Vertices of the image

a) (-5, -2)                                  (-5 + 6, -2 + 1) = (1, -1)

b) (-5, -3)                                  (-5 + 6, -3 + 1) = (1, -2)

c) (-4, -3)                                   (-4 + 6, -3 + 1) = (2, -2)

d) (-4, -5)                                  (-4 + 6, -5 + 1) = (2, -4)

e) (-3, -2)                                  (-3 + 6, -2 + 1) = (3, -1)

f) (-3, -4)                                   (-3 + 6, -4 + 1) = (3, -3)

g) (-2, -4)                                  (-2 + 6, -4 + 1) = (4, -3)

h) (-2, -5)                                  (-2 + 3, -5 + 1) = (1, -4)

<u>5. (x, y) → (x, y - 4)</u>

This is a translation four units down

Vertices of the preimage      Vertices of the image

a) (0, 2)                                    (0, 2 - 4) = (0, -2)

b) (0,1)                                      (0, 1 - 4) = (0, -3)

c) (2, 2)                                     (2, 2 - 4) = (2, -2)

d) (2,0)                                     (2, 0 - 4) = (2, -4)

<u>6. (x, y) → (x - 1, y + 4)</u>

This is a translation one unit to the left and four units up.

Vertices of the pre-image     Vertices of the image

a) (-4, -1)                                   (-4 - 1, -1 + 4) = (-5, 3)

b) (-4 - 5)                                  (-4 - 1, -5 + 4) = (-5, -1)

c) (-2, -4)                                  (- 2 - 1, -4 + 4) = (-3, 0)

d) (-2, -5)                                 (-2 - 1, -5 + 4) = (-3, -1)

8 0
3 years ago
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