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sveticcg [70]
3 years ago
14

(−4 + 5) − (6 + 7) = x x 0

Mathematics
1 answer:
weeeeeb [17]3 years ago
3 0

1

Add the numbers

(

−

4

+

5

)

−

(

6

+

7

)

=

0

({\color{#c92786}{-4}}+{\color{#c92786}{5}})-(6+7)=xx^{0}

(−4+5)−(6+7)=xx0

(

1

)

−

(

6

+

7

)

=

0

({\color{#c92786}{1}})-(6+7)=xx^{0}

(1)−(6+7)=xx0

2

Add the numbers

1

−

(

6

+

7

)

=

0

1-({\color{#c92786}{6}}+{\color{#c92786}{7}})=xx^{0}

1−(6+7)=xx0

1

−

(

1

3

)

=

0

1-({\color{#c92786}{13}})=xx^{0}

1−(13)=xx0

3

Multiply the numbers

1

−

1

⋅

1

3

=

0

1{\color{#c92786}{-1}} \cdot {\color{#c92786}{13}}=xx^{0}

1−1⋅13=xx0

1

−

1

3

=

0

1{\color{#c92786}{-13}}=xx^{0}

1−13=xx0

4

Subtract the numbers

1

−

1

3

=

0

{\color{#c92786}{1-13}}=xx^{0}

1−13=xx0

−

1

2

=

0

{\color{#c92786}{-12}}=xx^{0}

−12=xx0

5

Combine exponents

−

1

2

=

0

-12={\color{#c92786}{xx^{0}}}

−12=xx0

−

1

2

=

1

-12={\color{#c92786}{x^{1}}}

−12=x1

Show less

Solution

−

1

2

=

1

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Answer:  The answers are given below.

Step-by-step explanation:  We are given to factorise some quadratic expressions.

We will be using the following factorisation formula:

x^2+ax+bx+ab=(x+a)(x+b).

The factorisation is as follows:

We have

(1)~3ab-5ax+9b-15x\\\\=a(3b-5x)+3(3b-5x)\\\\=(3b-5x)(a+3).

Thus, option (B) (3b-5x)(a+3) is correct.

(2)~5x(x+3)+6(x+3)\\\\=(x+3)(5x+6).

Thus, option (B) (x+3)(5x+6) is correct.

(3)~2cb+7cx+2b+7x\\\\=c(2b+7x)+1(2b+7x)\\\\=(2b+7x)(c+1).

Thus, option (B) (2b+7x)(c+1) is correct

(4)~21x^3+35x^2+9x+15\\\\=7x^2(3x+5)+3(3x+5)\\\\=(3x+5)(7x^2+3).

Thus, option (D) (3x+5)(7x^2+3) is correct.

(5)~9xy+36x-10y-40\\\\=9x(y+4)-10(y+4)=(y+4)(9x-10).

Thus, option (C) (y+4)(9x-10) is the correct option.

(6)~14bx^2-7x^3-4b+2x\\\\=7x^2(2b-x)-2(2b-x)\\\\=(2b-x)(7x^2-2).

Thus, option (B) (2b-x)(7x^2-2) is the correct option.

(7)~4xy+9x+24y+54\\\\=x(4y+9)+6(4y+9)\\\\=(4y+9)(x+6).

Thus, option (D) (4y+9)(x+6)  is correct.

4 0
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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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