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3 years ago

Which ordered pairs are solutions to the inequality 4x+y>−6?

Mathematics

2 answers:

lisov135 [29]3 years ago

4 0

Answer:

(2,0) , (4,-20) and (-1,-1)

Step-by-step explanation:

$4x+y>-6$

WE need to select an ordered pair that is solution to our inequality

Lets check with each option

(2,0) , plug in 2 for x and 0 for y

$4x+y>-6$

$4(2)+0>-6$

$8>-6$ True

(−3, 6), plug in -3 for x and 6 for y

$4x+y>-6$

$4(-3)+6>-6$

$-6>-6$ False

(4, −20) , plug in 4 for x and -20 for y

$4x+y>-6$

$4(4)-20>-6$

$-4>-6$ True

(0, −9) , plug in 0 for x and -9 for y

$4x+y>-6$

$4(0)-9>-6$

$-9>-6$ False

(-1, −1) , plug in -1 for x and -1 for y

$4x+y>-6$

$4(-1)-1>-6$

$-5>-6$ True

Bingel [31]3 years ago

3 0

2,0 solution

-3,6no solution

4,-20 no solution

0,-9 no solution

-1,-1 solution

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3 years ago

2.) Combine like terms and simplify the expression: 2x + 5y+ 7x Your answer

SashulF [63]

Answer:

9x + 5y

Step-by-step explanation:

Here, in 2x + 5y+ 7x, the 'like terms' are 2x and 7x, and the sum of these terms is 9x. Thus, 2x + 5y+ 7x becomes 9x + 5y.

5 0

3 years ago

The value of 4 less than the product of 0.25 and x is greater than 6.

ella [17]

Answer:

0.25x - 4 > 6

7 0

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 IMAGE 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.

1. (x, y) → (x + 3, y - 2)

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)

2. (x,y) → (x - 3, y + 5)

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)

3. (x, y) → (x + 4, y)

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)

4. (x, y) → (x + 6, y + 1)

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)

5. (x, y) → (x, y - 4)

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)

6. (x, y) → (x - 1, y + 4)

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

In one area, the lowest angle of elevation of the sun in winter is 27.5°. Find the minimum distance x that a plant needing full

Alex73 [517]

Answer:

The minimum distance x that a plant needing full sun can be placed from a fence that is 5 feet high is 4.435 ft

Step-by-step explanation:

Here we have the lowest angle of elevation of the sun given as 27.5° and the height of the fence is 5 feet.

We will then find the position to place the plant where the suns rays can get to the base of the plant

Note that the fence is in between the sun and the plant, therefore we have

Height of fence = 5 ft.

Angle of location x from the fence = lowest angle of elevation of the sun, θ

This forms a right angled triangle with the fence as the height and the location of the plant as the base

Therefore, the length of the base is given as

Height × cos θ

= 5 ft × cos 27.5° = 4.435 ft

The plant should be placed at a location x = 4.435 ft from the fence.

8 0

3 years ago