divided 16 by 4 which is 4. then multiply 2 by 6, which is 12. then multiply 4 by 12. Remember to always do ( ) first. 4 by 12 is 48. Then do 48 divided by 6 which is 8. then add 8+4 which is 12.

4 0

3 years ago

Multiply. Check picture.

VashaNatasha [74]

The answer is $3x^4-13x^3-x^2-11x+6$ .

Solution:

Use algebraic identity: $a^m\times a^n=a^{m+n}$

For example: $x^2\times x=x^{2+1}=x^3$

Given expression $(x^2-5x+2)$ and $(3x^2+2x+3)$ .

To multiply these equations.

$(x^2-5x+2)\times(3x^2+2x+3)$

$=x^2(3x^2+2x+3)-5x(3x^2+2x+3)+2(3x^2+2x+3)$

$=(3x^4+2x^3+3x^2)+(-15x^3-10x^2-15x)+(6x^2+4x+6)$

$=3x^4+2x^3+3x^2-15x^3-10x^2-15x+6x^2+4x+6$

Combine like terms together.

$=3x^4+(2x^3-15x^3)+(3x^2-10x^2+6x^2)-15x+4x+6$

$=3x^4-13x^3-x^2-11x+6$

$(x^2-5x+2)\times(3x^2+2x+3)=3x^4-13x^3-x^2-11x+6$

Hence the answer is $3x^4-13x^3-x^2-11x+6$ .

3 0

3 years ago

EasyElectrics supermarket orders light bulbs from suppliers, AA Electronics and AAA Electronics. EasyElectrics purchases 30% of

saw5 [17]

Answer: (a) 0.006

(b) 0.027

Step-by-step explanation:

Given : P(AA) = 0.3 and P(AAA) = 0.70

Let event that a bulb is defective be denoted by D and not defective be D';

Conditional probabilities given are :

P(D/AA) = 0.02 and P(D/AAA) = 0.03

Thus P(D'/AA) = 1 - 0.02 = 0.98

and P(D'/AAA) = 1 - 0.03 = 0.97

(a) P(bulb from AA and defective) = P ( AA and D)

= P(AA) x P(D/AA)

= 0.3 x 0.02 = 0.006

(b) P(Defective) = P(from AA and defective) + P( from AAA and defective)

= P(AA) x P(D/AA) + P(AAA) x P(D/AAA)

= 0.3(0.02) + 0.70(0.03)

= 0.027

3 0

3 years ago