Answer: $454.55
Step-by-step explanation:
Tito's weekly salary is always $ 500 per week.
But your weekly commission income depends on the sales you make. Your commissions are 55%.
If we call x the dollar amount of all sales made by Tito in a week, then his income z(x) is given by the following equation:
We wish that 
.
Then we equate the equation to 750 and solve for the variable x.
Total sales made by Tito must be $454.55
Volume of a pyramid = 
Given base is = 140 m
Given height is = 96 m
Volume = 
Hence, volume of pyramid is 4480 m³
First do ,60 divided by 7 =_____. Then do The product of that equation divided by 8. Then ur answer to that equation is a factor for the last equation u have to do witch is The product of ______ divided by 8, Then multiple that product by 3 like this (3❌_______=_______) then u have ur answer. Hope u understand more now that I've helped
Answer:
x² + 2x + [3\x - 1]
Step-by-step explanation:
Since the divisor is in the form of <em>x - c</em>, use what is called <em>Synthetic Division</em>. Remember, in this formula, -c gives you the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:
1| 1 1 -2 3
↓ 1 2 0
------------------
1 2 0 3 → x² + 2x + [3\x - 1]
You start by placing the <em>c</em> in the top left corner, then list all the coefficients of your dividend [x² + 5x - 36]. You bring down the original term closest to <em>c</em> then begin your multiplication. Now depending on what symbol your result is tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have no remainder, which in this case is a 3, so what you is set the divisor underneath the remainder of 3. Finally, your quotient is one degree less than your dividend, so that 1 in your quotient can be an x², 2 becomes <em>2x</em><em>,</em><em> </em>and the remainder of 3 is set over the divisor, giving you the other factor of <em>x² + 2x + [3\x - 1]</em>.
If you are ever in need of assistance, do not hesitate to let me know by subscribing to my channel [USERNAME: MATHEMATICS WIZARD], and as always, I am joyous to assist anyone at any time.
