K/3 + 4 -2k = -9k
First you add 9k to both sides
<span>k/3 + 4 -2k = -9k
</span> +9k +9k
<span>k/3 + 4 + 7k = 0
</span>
Now subtract 4 from both sides
k/3 + 4 + 7<span>k = 0
</span> -4 -4
k/3 + 7<span>k = -4
</span>
Now multiply 3 by both sides
k/3 + 7<span>k = -4
</span>x3 x3 x3
k +21k = -12
Now add k + 21k
22k = -12
Now divide both sides by 22
<span>22k = -12
</span>----- -----
22 22
k = -6/11
The answer is 2, 4, 8 and 16 (added=$30)
Take $14 from the highest amount = $2 for that envelope
Double the original $2 = $4
Double the $4 = $8
Double the $6 = $16
Thus, $2,4,8,16, but all in different envelopes
1+tan^2(A) = sec^2(A) [Pythagorean Identities]
tan^2(A)cot(A) = tan(A)[tan(A)cot(A)] = tan(A)[1] = tan(A)
*see photo for complete solution*
Answer:
I am pretty sure the answer is 30 blocks :)
4 + 11 + 6 + 9 = 30 blocks