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Nikolay [14]
3 years ago
6

96 13=13 96 what property

Mathematics
2 answers:
andrew11 [14]3 years ago
8 0
Commutative property
tatiyna3 years ago
6 0

Answer: Commutative property of multiplication.


Step-by-step explanation:

96\cdot13=13\cdot96=1248

In algebra, the commutative property of multiplication tells that two numbers can be multiplied in either order.

ab=ba, where a and b be any two real numbers

It doesn't matter that which number should be written first or next, because if you change the order of product, the answer remains the same. The answer doesn't change with the change of order.

For example :- 3×5=5×3=15

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The scale is 11 feet per one inch.

Step-by-step explanation:

\frac{165feet}{15inches} = 11feet / inch

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1) Determine the discriminant of the 2nd degree equation below:
Aleksandr-060686 [28]

\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}

We have, Discriminant formula for finding roots:

\large{ \boxed{ \rm{x =  \frac{  - b \pm \:  \sqrt{ {b}^{2}  - 4ac} }{2a} }}}

Here,

  • x is the root of the equation.
  • a is the coefficient of x^2
  • b is the coefficient of x
  • c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 5\pm  \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 5  \pm  \sqrt{25 - 24} }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 5 \pm 1}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x =  - 2 \: or  - 3}}}

❒ p(x) = x^2 + 2x + 1 = 0

\large{ \rm{ \longrightarrow \: x =  \dfrac{  - 2 \pm  \sqrt{ {2}^{2}  - 4 \times 1 \times 1} }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - 2 \pm 0}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x =  - 1 \: or \:  - 1}}}

❒ p(x) = x^2 - x - 20 = 0

\large{ \rm{ \longrightarrow \: x =  \dfrac{ - ( - 1) \pm  \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{1 \pm 9}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \:  - 4}}}

❒ p(x) = x^2 - 3x - 4 = 0

\large{ \rm{ \longrightarrow \: x =   \dfrac{  - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{3 \pm \sqrt{9  + 16} }{2 \times 1} }}

\large{ \rm{ \longrightarrow \: x =  \dfrac{3  \pm 5}{2} }}

So here,

\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \:  - 1}}}

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