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LiRa [457]
3 years ago
11

Factorize 9(a+1)^2-81​

Mathematics
1 answer:
lora16 [44]3 years ago
6 0

Answer:

The answer is 9(a-2)(a+4).

Step-by-step explanation:

Use a^{2} -b^{2} =(a-b)(a+b)  

9(a+1)^{2} -81= 9( (a+1)^{2}-9)=  9( (a+1)^{2}-3^{2} )= 9(a+1-3)(a+1+3)\\=9(a-2)(a+4).

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Answer:

C. g(x) = 4x²  

Step-by-step explanation:

The general equation for a parabola is  

y = ax² + bx +c  

Since ƒ(x) = x², a=1, b =0, c =0

For g(x), the vertex is still at  the origin, so

g(x) = ax²

The graph passes through (1,4).

Insert the coordinates of the point.

4 = a(1)²

a = 4

g(x) = 4x²

The figure below shows that the graph of g(x) = 4x² passes through the point (1, 4).

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Identifying and Graphing Sequences: lesson 4-1
goldfiish [28.3K]

Answer:

1) Domain: {   1,2,3,4,5,6}

Range: {12,24,36,48,60,72}

2) First four terms are: 2,5,8,11

3) First four terms are: 8,13,20,29

4) First four terms are: 0,0,2,6

5) First four terms are: 0,1,1.41,1.73

Step-by-step explanation:

1.

n       1      2     3     4      5      6

f(n)    12    24    36   48    60    72

Domain: {   1,2,3,4,5,6}

Range: {12,24,36,48,60,72}

Write first four terms of each sequence

2) f(n)=3n-1

Put n = 1, 3(1)-1 = 3-1 = 2

Put n = 2, 3(2)-1 = 6-1 = 5

Put n = 3, 3(3)-1 = 9-1 = 8

Put n = 4, 3(4)-1 = 12-1 = 11

So, First four terms are: 2,5,8,11

3) f(n)=n^2+2n+5

Put n = 1, (1)^2+2(1)+5= 1+2+5 = 8

Put n = 2, (2)^2+2(2)+5 = 4+4+5= 13

Put n = 3, (3)^2+2(3)+5 = 9+6+5 = 20

Put n = 4, (4)^2+2(4)+5 = 16+8+5 = 29

So, First four terms are: 8,13,20,29

4) f(n)=(n-1)(n-2)

Put n = 1, (1-1)(1-2) = 0(-1) = 0

Put n = 2, (2-1)(2-2) = 1(0) = 0

Put n = 3, (3-1)(3-2) = 2(1) = 2

Put n = 4, (4-1)(4-2) = 3(2) = 6

So, First four terms are: 0,0,2,6

5) f(n) = \sqrt{n-1}

Put n=1 \sqrt{1-1}=\sqrt{0}=0

Put n=2 \sqrt{2-1}=\sqrt{1}=1

Put n=3 \sqrt{3-1}=\sqrt{2}= 1.41

Put n=4 \sqrt{4-1}=\sqrt{3}=1.73

So, First four terms are: 0,1,1.41,1.73

8 0
3 years ago
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