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Lemur [1.5K]
3 years ago
7

What is the factorization of 121b^4-49

Mathematics
1 answer:
Nezavi [6.7K]3 years ago
4 0

121^4-49

(11^2+7)(11^2-7) 1. 121. 49

11. × 11. 7 × 7

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Answer:

88% of the free throws were made

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What is the equation when y=5x+4 is reflected over the x-axis, y-axis & y=x?
tigry1 [53]

Answer:

A) y=-5x-4

B) y=-5x+4

C) y=\frac{x-4}{5}

Step-by-step explanation:

So we have the equation:

y=5x+4

Let's write this in function notation. Thus:

y=f(x)=5x+4

A)

To flip a function over the x-axis, multiply the function by -1. Thus:

f(x)=5x+4\\-(f(x))=-(5x+4)

Simplify:

-f(x)=-5x-4

B) To flip a function over the y-axis, change the variable x to -x. Thus:

f(x)=5x+4\\f(-x)=5(-x)+4

Simplify:

f(-x)=-5x+4

C) A reflection over the line y=x is synonymous with finding the inverse of the function.

To find the inverse, switch x and f(x) and solve for f(x):

f(x)=5x+4

Switch:

x=5f^{-1}(x)+4

Subtract 4 from both sides:

x-4=5f^{-1}(x)

Divide both sides by 5:

f^{-1}(x)=\frac{x-4}{5}

And we're done :)

7 0
3 years ago
2. Check the boxes for the following sets that are closed under the given
son4ous [18]

The properties of the mathematical sequence allow us to find that the recurrence term is 1 and the operation for each sequence is

   a) Subtraction

   b) Addition

   c) AdditionSum

   d) in this case we have two possibilities

       * If we move to the right the addition

       * If we move to the left the subtraction

The sequence is a set of elements arranged one after another related by some mathematical relationship. The elements of the sequence are called terms.

The sequences shown can be defined by recurrence relations.

Let's analyze each sequence shown, the ellipsis indicates where the sequence advances.

a) ... -7, -6, -5, -4, -3

We can observe that each term has a difference of one unit; if we subtract 1 from the term to the right, we obtain the following term

        -3 -1 = -4

        -4 -1 = -5

        -7 -1 = -8

Therefore the mathematical operation is the subtraction.

b) 0. \sqrt{1}. \sqrt{4}, \sqrt{9}, \sqrt{16}, \sqrt{25}  ...

In this case we can see more clearly the sequence when writing in this way

      0, \sqrt{1^2}. \sqrt{2^2}, \sqrt{3^2 } . \sqrt{4^2} , \sqrt{5^2}

each term is found by adding 1 to the current term,

      \sqrt{(0+1)^2} = \sqrt{1^2} \\\sqrt{(1+1)^2} = \sqrt{2^2}\\\sqrt{(2+1)^2} = \sqrt{3^2}\\\sqrt{(5+1)^2} = \sqrt{6^2}

Therefore the mathematical operation is the addition

c)   ... \frac{-10}{2}. \frac{-8}{2}, \frac{-6}{2}, \frac{-4}{2}. \frac{-2}{2}. ...

      The recurrence term is unity, with the fact that the sequence extends to the right and to the left the operation is

  • To move to the right add 1

           -\frac{-10}{2} + 1 = \frac{-10}{2}  -   \frac{2}{2}  = \frac{-8}{2}\\\frac{-8}{2} + \frac{2}{2} = \frac{-6}{2}

  • To move left subtract 1

         \frac{-2}{2} - 1 = \frac{-4}{2}\\\frac{-4}{2} - \frac{2}{2} = \frac{-6}{2}

         

Using the properties the mathematical sequence we find that the recurrence term is 1 and the operation for each sequence is

   a) Subtraction

   b) Sum

   c) Sum

   d) This case we have two possibilities

  •  If we move to the right the sum
  •  If we move to the left we subtract

Learn more here: brainly.com/question/4626313

5 0
3 years ago
I need help with this​
VLD [36.1K]

Answer:

im sorry wish i could

Step-by-step explanation:

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(2x)2 > 3x2 is true the first two you can not have two x’s on the same side because where would you put them both when solving them the last one I’m not sure about
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