LN = 120 and a = 10
<u>Step-by-step explanation:</u>
If we need to find the variable a as, LN is the line and M is the mid point between L and N. So LN = LM + MN
LM = 4a = 40
a = 
= 10
MN = 8a = 8×10 = 80
So LN = LM + MN
LN = 40 + 80 = 120
Answer:
A. 92% bc 100-8= 92 so 216 animals are 92% of the now 233 animals after 8% increase.
B. 100% bc that is the final # of animals.
\left[a _{3}\right] = \left[ \frac{ - b^{2}}{6}+\frac{\frac{ - b^{4}}{3}+\left( \frac{-1}{3}\,i \right) \,\sqrt{3}\,b^{4}}{2^{\frac{2}{3}}\,\sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}+\frac{\frac{ - \sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}{24}+\left( \frac{1}{24}\,i \right) \,\sqrt{3}\,\sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}{\sqrt[3]{2}}\right][a3]=⎣⎢⎢⎢⎢⎡6−b2+2323√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))3−b4+(3−1i)√3b4+3√224−3√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))+(241i)√33√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))⎦⎥⎥⎥⎥⎤
Step-by-step explanation:
-(-5x + 3) = 17
5x - 3 = 17
5x = 20
x = 4.
1. x = 2
2. x = 9
3. a = 3
4. x = 13
5. c = 3
6. s = 3
7. x = 7
8. c = 0
9. b = 1
10. c = 4
11. x = 4
12. x = 8
13. x = 12
14. y = 10
15. x = 11
16. x = 10
17. x = 6
18. x = 11
