<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
To solve this, you need to divide 127.5 by 51:
127.5/51 = 2.5
To find the percent, multiply 2.5 by 100:
2.5 x 100% = 250%
So 127.5 is 250% of 51
The answer is given by the empirical rule, 68%.
Answer:
26 bags
Step-by-step explanation:
A bag of cat chows = 30 lbs
Amount of cat chows needed in a week = 15 lbs = ½ bag of cat chow
There are at least 52 weeks in a year.
The least number of cat chows bags needed at the shelter in 1 year = 52 weeks × ½ bag of chow
= 52 × ½
= 52/2
= 26 bags of cat chow
At least, 26 bags would be needed in 1 year (52 weeks) at this wild life shelter if 15 lbs is consumed weekly.
