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MakcuM [25]
2 years ago
13

Is there anyone that can help me with this problem?

Mathematics
2 answers:
Flura [38]2 years ago
8 0

Answer

Step-by-step explanation:

so, this triangle is an "impossible' triangle, b/c  even when the 5 and 4 sides are laid out, in opposite directions, they only make a distance of 9. so the side of 12 is not ever going to work in this triangle, it's just too long.  so the answer is none of the choices are correct.  To be fair, this question is incorrect.  The picture that is drawn, makes it look like the triangle could be real. when it's not.  like an Escher drawing. if  you don't know Escher drawings look  them up                                                                                        

saul85 [17]2 years ago
6 0

Answer:

your answer will be option 2

None of the choices are correct.

Step-by-step explanation:

have a nice day and good luck!!!

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60 pts!!! HELP Find the inverse of the function f(x)= 4/9 x-4<br> show work
earnstyle [38]

Answer:

Inverse Function: f^-1(x) = 9/4x + 9

Step-by-step explanation:

To find  the inverse function we can interchange the position of the variables x and y, and then solve for y;

y = 4/9x - 4 => x = 4/9y - 4,

x = 4/9y - 4,

4/9y = x + 4,

y = 9/4x + 9/4(4),

y = 9/4x + 9/ f^-1(x) = 9/4x + 9

6 0
3 years ago
Victor buys some six-packs of soda for a party. He buys 42 cans in all. How many six-packs of soda did Victor buy?​
Whitepunk [10]

Answer: 7

Step-by-step explanation: 7*6=42

5 0
3 years ago
Read 2 more answers
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
Please help with this question please
puteri [66]

Answer:

Choice A

Step-by-step explanation:

(4m^5n^2/m^2n)^3

dividing exponents = subtraction

(4m^3n)^3

4^3 = 64

(m^3 )^3 = m^3x3 = m^9

64m^9n^3

5 0
3 years ago
Please help will mark brainliest if correct
Annette [7]

Answer:

umm

Step-by-step explanation:

u fogot to send the pic

7 0
3 years ago
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