Please help! algebra 2

c

Step-by-step explanation:

Let's go one by one.

Reminder before we start anything: i is an imaginary number that equals .

Numerator first:

(5 + i)(6 - 5i)

Your first instinct should be to multiply it out to try to get rid of the imaginary numbers, so FOIL:

(5)(6) + (5)(-5i) + (i)(6) + (i)(-5i)

30 - 25i + 6i - 5i²

30 - 19i + 5

35 - 19i

Alright, now the denominator; we just have to distribute:

2i(-3 + 7i)

-6i + 14i²

-14 - 6i

Now we put the numerator and denominator together to get .

Unfortunately this doesn't look like any of our answer choices yet, so we have to multiply the top and bottom of the fraction by the conjugate of the denominator to get a real number on the bottom.

What's the conjugate? Well the bottom is -14 - 6i so the conjugate is -14 + 6i.

× = c

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl