just divide 200 until u get 25 i think but idk the second one
Answer:
If `r` and `R` and the respective radii of the smaller and the bigger semi-circles then the area of the shaded portion in the given figure is: (FIGURE) `pir^2\ s qdotu n i t s` (b) `piR^2-pir^2\ s qdotu n i t s` (c) `piR^2+pir^2\ s qdotu n i t s` (d) `piR^2\ s qdotu n i t s`
Step-by-step explanation:
Answer:

Step-by-step explanation:
Given the rational expression:
, to express this in simplified form, we would need to apply the concept of partial fraction.
Step 1: factorise the denominator





Thus, we now have: 
Step 2: Apply the concept of Partial Fraction
Let,
= 
Multiply both sides by (x - 1)(x - 4)
= 

Step 3:
Substituting x = 4 in 





Substituting x = 1 in 





Step 4: Plug in the values of A and B into the original equation in step 2


Answer:
A. 2x(x+1)(x-6); 0, -1, 6
Step-by-step explanation:
The zeros are the values of x that make the factors zero. That is, for binomial factors, they are the opposite of the constant in the binomial factor. For example, the factor (x+1) will be zero when x = -1, so that -1+1 = 0.
This observation eliminates choices B and C.
__
The product of binomial factors looks like this:
(x +a)(x +b) = x² +(a+b)x +ab . . . . . x-coefficient is (a+b)
Once 2x is factored from the given polynomial, the resulting quadratic is ...
x^2 -5x -6
This means the sum of the constants in the binomial terms must be -5. That will only be the case for choice A.
Answer:
Your brother can buy anywhere from 1 to 6 baseball cards depending on how rare they are!