<h3>
Answer:</h3>
- A) p = 5, one solution
- B) no solutions
- C) infinite solutions
<h3>
Step-by-step explanation:</h3>
A) Add 19-5p to each side of the equation:
... 10 = 2p
... 5 = p . . . . . divide by the coefficient of p
B) Subtract 5p from both sides of the equation:
... -9 = -19 . . . . . there is <em>no value of p</em> that will make this true. (No solution.)
C) Subtract 5p from both sides of the equation:
... -9 = -9 . . . . . this is true for <em>every value of p</em>. (Infinite solutions.)
There is a multiple zero at 0 (which means that it touches there), and there are single zeros at -2 and 2 (which means that they cross). There is also 2 imaginary zeros at i and -i.
You can find this by factoring. Start by pulling out the greatest common factor, which in this case is -x^2.
-x^6 + 3x^4 + 4x^2
-x^2(x^4 - 3x^2 - 4)
Now we can factor the inside of the parenthesis. You do this by finding factors of the last number that add up to the middle number.
-x^2(x^4 - 3x^2 - 4)
-x^2(x^2 - 4)(x^2 + 1)
Now we can use the factors of two perfect squares rule to factor the middle parenthesis.
-x^2(x^2 - 4)(x^2 + 1)
-x^2(x - 2)(x + 2)(x^2 + 1)
We would also want to split the term in the front.
-x^2(x - 2)(x + 2)(x^2 + 1)
(x)(-x)(x - 2)(x + 2)(x^2 + 1)
Now we would set each portion equal to 0 and solve.
First root
x = 0 ---> no work needed
Second root
-x = 0 ---> divide by -1
x = 0
Third root
x - 2 = 0
x = 2
Forth root
x + 2 = 0
x = -2
Fifth and Sixth roots
x^2 + 1 = 0
x^2 = -1
x = +/- 
x = +/- i
Answer:
81 9/10
Step-by-step explanation:
Actually, this answer would be true. Why?
The first equation is: a(sub <em>n</em>) = 8, 13, 18, 23
The second is: a(sub 1)=8 ; a(sub <em>n</em>)= a(sub <em>n</em>-1)+5
if you wish to find the second term, plug two into the equation for <em /><em>n</em>
8+5=13
to find the third, plug the second term, 13, in for <em>n.</em>
13+5=18.
Hope this helped! I know it's a bit on the late side, but at least you can get the general idea!
Answer:

Step-by-step explanation:
* Look to the attached file