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MrRissso [65]
3 years ago
11

I need help with this please its due today :(

Mathematics
1 answer:
yKpoI14uk [10]3 years ago
4 0

Answer:

a) \ 6^2 \times 6^3 = 6^{2+3} = 6^5 = 7776\\\\b)\ \frac{(-3)^5 }{(-3)^2} = (-3)^{5-2} = (-3)^3 = -27\\\\c)\ [(-2)^3]^4 = (-2)^{3\times 4} = (-2)^{12} = 4096 \\\\d)\ (\frac{1}{999})^0 = 1,  anything \ raise \ to \ zero = 1\\\\e) \ (\frac{4}{3})^{-2} = 4^{-2} \times 3^2 = \frac{9}{16}\\\\f) \ \frac{2^5 \times 2^9}{2^{16}} = 2^{5+9-16} = 2^{-2} = \frac{1}{4}

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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
A shopper keeper sold a total of 15 boxes of pencils on Monday and Tuesday. She still three more boxes on Monday then on Tuesday
Nookie1986 [14]
I think 180 that should be your answer sorry i did not get here quick enough

8 0
2 years ago
Given f(x)=x3 and g(x)= 1-5x2, fine (fog)(x) and it’s Domain
andre [41]

Answer:

Option B. f(g(x)) = (1-5x ^ 2) ^ 3  all real numbers

Step-by-step explanation:

We have

f(x) = x ^ 3 and g(x) = 1-5x ^ 2

They ask us to find

(fog)(x) and it's Domain

To solve this problem we must introduce the function g(x) within the function f(x)

That is, we must do f(g(x)).

So, we have:

f(x) = x ^ 3

g(x) = 1-5x ^ 2

Then:

f(g(x)) = (1-5x ^ 2) ^ 3

The domain of the function f(g(x)) is the range of the function g(x) = 1-5x ^ 2.

Since the domain and range of g(x) are all real numbers then the domain of f(g(x)) are all real numbers

Therefore the correct answer is the option b: f(g(x)) = (1-5x ^ 2) ^ 3

And his domain is all real.

5 0
3 years ago
Read 2 more answers
HELP ASAP!!!!!!!!!!
chubhunter [2.5K]

Answer:

7.5

Step-by-step explanation:

\sqrt{0\\.3} = 0.5477225575

Multiply it by 5

= 2.738612788

Square it

7.5 ft^{2}

Hope this helped. :)

5 0
3 years ago
X = 8,1,3,3,5
Ghella [55]

Answer:

2.600 = Exact

2.6 = Base Form

Step-by-step explanation:

?

6 0
2 years ago
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