Determine whether the two expressions are equivalent. Explain your reasoning.

1 answer:

6 0

Let us evaluate each pair of expressions one at a time.
8. (n²+4) - n² and 4n.
(n²+4) - n² = n²+ 4 - n² = 4 which is not equal to 4n.
NOT EQUIVALENT

9. 3x + 5 and 2(x + 3)
2(x + 3) = 2x + 6 which is not equal to 3x + 5.
NOT EQUIVALENT

10. 15 - 6x and 15(1 - 6x)
15(1 - 6x) = 15 - 90x which is not equal to 15 - 6x
NOT EQUVALENT

11. (y + y + 2 + y) + 3y and 6y + 2
(y +y +2 + y) + 3y = y+y+y +2 + 3y = 3y + 2 + 3y = 6y + 2.
It matches the other expression.
EQUIVALENT

12. 8y - 3 + 10y and 3(6 - 1)
8y - 3 + 10y = 8y + 10y - 3 = 18y - 3
3(6 - 1) = 3(5) = 15
The two expressions do not match.
NOT EQUIVALENT.

Answer: Only the expressions in number 11 are equivalent.

You might be interested in

Find the product of z1 and z2, where z1 = 7(cos 40° + i sin 40°) and z2 = 6(cos 145° + i sin 145°).

Oduvanchick [21]

For two complex numbers $z_1=re^{i\theta}=r(\cos\theta+i\sin\theta)$ and $z_2=se^{i\varphi}=s(\cos\varphi+i\sin\varphi)$ , the product is

$z_1z_2=rse^{i(\theta+\varphi)}=rs(\cos(\theta+\varphi)+i\sin(\theta+\varphi))$

That is, you multiply the moduli and add the arguments. You have $z_1=7e^{i40^\circ}$ and $z_2=6e^{i145^\circ}$ , so the product is

$z_1z_2=7\times6(\cos(40^\circ+145^\circ)+i\sin(40^\circ+145^\circ)=42(\cos185^\circ+i\sin185^\circ)=42e^{i185^\circ}$