Whether dividing constant terms or polynomials, we always have definitive terms when it comes to division. Suppose we say, 10x divided by 2. The dividend is the 10x and the divisor is the 2. In other words, the dividend is the number to be divided by the divisor, to obtain the answer called the quotient.
When dividing polynomials, your main goal is to be able to divide the dividend evenly into the <em>divisor</em>. For example, we divide x²+2x+1 by x+1. The first thing you're going to focus is, what term will completely divide the first term of the polynomial? That would be x. Why? Because when you multiply x with x+1, the product is x²+x. When you subtract this from the polynomial, the x² will cancel out. All you have to do is subtract x from 2x, yielding x. Then, you carry down the last term of the equation: +1. You do the steps again. The term that will completely divide x+1 by x+1 is 1. When you subtract the two, you will come up with zero. That means there is no remainder. The polynomial is divisible by the divisor.
x + 1
------------------------------------
x+1| x²+2x+1
- x²+x
----------------------
x +1
- x +
------------
0
Answer: P^2 - 10p + 24 = 0
Step-by-step explanation:
Given that a company’s weekly revenue, in thousands, is modeled by the equation
R = -p2 + 14p,
where p is the price of the product it makes. The company is considering hiring an outside source to distribute its products, which will cost the company 4p + 24 thousand dollars per week.
If the company want to break even, the cost of hiring distributors will be equal to the revenue per week.
Therefore,
-P^2 + 14p = 4p + 24
Collect the like terms
-P^2 + 14p - 4p - 24 = 0
-P^2 + 10p - 24 = 0
Multiply all by minus sign
P^2 - 10p + 24 = 0
75<span>°. So divide 120 by 8, then multiply 8 with 15. That's the measure of the larger angle. </span>
x/240 = 82/100
240 : 100 = 2.4
so 240 is divided with 2,4 giving 100, this will work with 82 too. So, 82 × 2.4 = 196,8
Answer : x = 196,8
prove :
196,8/240 = 82/100 = 41/50
see the picture
Answer:
y=t−1+ce
−t
where t=tanx.
Given, cos
2
x
dx
dy
+y=tanx
⇒
dx
dy
+ysec
2
x=tanxsec
2
x ....(1)
Here P=sec
2
x⇒∫PdP=∫sec
2
xdx=tanx
∴I.F.=e
tanx
Multiplying (1) by I.F. we get
e
tanx
dx
dy
+e
tanx
ysec
2
x=e
tanx
tanxsec
2
x
Integrating both sides, we get
ye
tanx
=∫e
tanx
.tanxsec
2
xdx
Put tanx=t⇒sec
2
xdx=dt
∴ye
t
=∫te
t
dt=e
t
(t−1)+c
⇒y=t−1+ce
−t
where t=tanx
