<h3>(-3j²k³)²(2j²k)³</h3>
(-3j²k³)²(2j²k)³ = <em>When a power is raised to a power the exponents have to be multiplied.</em>
= (-3²j⁽²*²⁾k⁽³*²⁾)(2³j⁽²*³⁾k³) = <em>We can take out the constants</em>
= (9)(8)(j⁴k⁶)(j⁶k³) = <em>We can group the same variables</em>
= 72(j⁴j⁶)(k⁶k³) = <em>When multiplying two powers that have the same base, you have to add the exponents.</em>
= 72 j⁽⁴⁺⁶⁾k⁽⁶⁺³⁾ = 72j¹⁰k⁹
Answer = 72j¹⁰k⁹
Hope this helps!
Answer:
Yes they are congruent by SAS (wrong answer)
Edit: the correct answer is by HL
9514 1404 393
Answer:
C. 3x² +24
Step-by-step explanation:
Use the given function definitions and simplify.

5 (2.3) = 11.5
Hope this helps
(6÷2)+(5+2×3)=24
I got this so this is the answer
;D