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3 years ago

Select the postulate that is illustrated for the real numbers.

Mathematics

2 answers:

VLD [36.1K]3 years ago

8 0

Answer: The correct option is

(1) The commutative postulate for multiplication

Step-by-step explanation: We are given to select the correct postulate that is illustrated for the real numbers as follows :

6 . 12 = 12 . 6.

We know that

if a and b are any two real numbers, then the equality a . b = b . a illustrates the commutative postulate for multiplication.

So, if a = 6 and b = 12, then we get

6 . 12 = 12 . 6.

Thus, the illustrated postulate is the commutative postulate for multiplication.

Option (1) is CORRECT.

jeka57 [31]3 years ago

6 0

The answer would be 1. The commutative postulate for multiplication. Hope this helped! Good luck! :)

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In a triangle ABC, measure of angle B is 90 degrees. AB is 3x-2 units and BC is x+3. If the area of the triangle is 17 sq cm, fo

Scorpion4ik [409]

Answer:

$x=\frac{8}{3}\ cm$

Step-by-step explanation:

we know that

The area of the right triangle ABC is equal to

$A=\frac{1}{2}(AB)(BC)$

we have

$A=17\ cm^2$

$AB=(3x-2)\ cm$

$BC=(x+3)\ cm$

substitute the values

$17=\frac{1}{2}(3x-2)(x+3)$

$34=(3x-2)(x+3)$

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$3x^2+7x-40=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$3x^2+7x-40=0$

$a=3\\b=7\\c=-40$

substitute in the formula

$x=\frac{-7(+/-)\sqrt{7^{2}-4(3)(-40)}} {2(3)}$

$x=\frac{-7(+/-)\sqrt{529}} {6}$

$x=\frac{-7(+/-)23} {6}$

$x=\frac{-7(+)23} {6}=\frac{16}{6}=\frac{8}{3}$

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therefore

The solution is

$x=\frac{8}{3}\ cm$

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Harman [31]

Answer:

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3 years ago