<span>Equation at the end of step 1 :</span><span> (((x3)•y)-(((3x2•y6)•x)•y))-6y = 0
</span><span>Step 2 :</span><span>Step 3 :</span>Pulling out like terms :
<span> 3.1 </span> Pull out like factors :
<span> -3x3y7 + x3y - 6y</span> = <span> -y • (3x3y6 - x3 + 6)</span>
Trying to factor a multi variable polynomial :
<span> 3.2 </span> Factoring <span> 3x3y6 - x3 + 6</span>
Try to factor this multi-variable trinomial using trial and error<span>
</span>Factorization fails
<span>Equation at the end of step 3 :</span><span> -y • (3x3y6 - x3 + 6) = 0
</span><span>Step 4 :</span>Theory - Roots of a product :
<span> 4.1 </span> A product of several terms equals zero.<span>
</span>When a product of two or more terms equals zero, then at least one of the terms must be zero.<span>
</span>We shall now solve each term = 0 separately<span>
</span>In other words, we are going to solve as many equations as there are terms in the product<span>
</span>Any solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation :
<span> 4.2 </span> Solve : -y = 0<span>
</span>Multiply both sides of the equation by (-1) : y = 0
Answer:
No i do for other stuff but not this and the answer is 18
Step-by-step explanation:
We are asked to find 5% od 320, because we know that 5% of 320 insurances are rejected.
5%=0.05
Now we can multiply 0.05 by 320
0.05*320=16 - its our result
Answer:
Step-by-step explanation:
GIVEN: two two-letter passwords can be formed from the letters A, B, C, D, E, F, G and H.
TO FIND: How many different two two-letter passwords can be formed if no repetition of letters is allowed.
SOLUTION:
Total number of different letters 
for two two-letter passwords 
different are required.
Number of ways of selecting different letters from 
letters
Hence there are different two-letter passwords can be formed using letters.
