divide -5/0.......undefined is the answer....right........please help?

Question

add 13+(-11) i say (-2.)...........
multiply (5a-3b)^2 i say (25a^2+9b^2)....
divide x^4+x^2-80/x+3 = ............can't seem to ge this........please help someone......

Answer 1

(A) -5 / 0 is indeed undefined.

(B) 13 + (-11) = 2. You have a greater value of positives added to a lesser value of negatives. If you have $13 and owe someone $11 [13 + (-11)], after you pay it, you still have $2 for yourself. It's a positive amount.

(C) (5a - 3b)² = 25a² - 30ab + 9b². Missing the middle term of a squared binomial is a very common mistake.

Think of it as (5a - 3b)(5a - 3b), then use your "FOIL" method.
25a² - 15ab - 15ab + 9b², then combine like terms.

(D) Your division problem is written ambiguously, but I'm guessing you mean (x^4 + x² - 80) / (x + 3)?
Use long division to get this, if you haven't yet learned synthetic division. Do this one term at a time.

You need x³ times x to get x^4.
(x³ + __ + __ + __) * (x + 3) = x^4 + 3x³
You need to get rid of the 3x³.
(x³ - 3x² + __ + __) * (x + 3) = (x^4 + 3x³) + (-3x³ + 9x²) = x^4 + 9x²
You're trying to get x², not 9x², so you have to get rid of 8x².
(x³ - 3x² - 8x + __) * (x + 3) = (x^4 + 9x²) + (-8x² - 24x) = x^4 + x² - 24x
You need to get rid of the -24x.
(x³ - 3x² - 8x + 24) * (x + 3) = (x^4 + x² - 24x) + (24x + 72) = x^4 + x² + 72
You have 152 too many units [72 - (-80) = 152], so use a remainder of -152.

[x³ - 3x² - 8x + 24 - (152)/(x + 3)] * (x + 3) = (x^4 + x² + 72) - 72 = x^4 + x² - 80.

Answer 2

-5 divided by 0 is undefined. Anything divided by 0 is undefined.

13 + (-11) = 2 Imagine going 13 steps up a ladder and then 11 steps down. You would still be two steps above the ground, so the answer is +2.

(5a-3b)^2 = 25a^2 - 30ab + 9b^2
Imagine a=10 and b=1. 47 sq is not 2509, which is what (25a^2 + 9b^2) comes to.

when you multiply 5a -3b times 5a-3b, you multiply each of the terms by each of the other. That's 5a * 5a + 5a*(-3b) + (-3b)*5a + (-3b)*(-3b).

Answer 3

5/0 is undefined. If 5/0 is some number "q", bad things happen...

5/0 = q (multiply both sides by zero!)

5 = 0 x q which is bad, because zero multiplied by any number should be zero, not five!

(5a-3b)^2 = 25a^2 - 30ab + 9b^2. Multiply (5a-3b)(5a-3b) out, and you'll see.

The division problem requires polynomial long division, and is too long for me to explain here. But the link below gets you started in the right place.

Answer 4

Any number (other than 0) divided by 0 has no solution. Only 0/0 is undefined.

13 + (-11) is the same as 13 - 11, or 2. Not -2.

(5a - 3b)^2 is the same as (5a - 3b)(5a - 3b). You have to use FOIL (first, outside, inside, last) to multiply, you can't just square each term.
(5a - 3b)(5a -3b)
= 25a^2 -15ab - 15ab + 9b^2
= 25a^2 -30ab + 9b^2

Answer 5

-5/0 = infinity, or undefined
13+(-11) = 13 - 11 = 2
(5a-3b)^2 = 25a^2 + 9b^2 -30ab
the last one you need to do long division

Answer 6

well.. first of all... 13 + (-11) = 2.. not -2
(5a-3b)^2 is 25a^2-30ab+9b^2.. you totally left out the -30ab
ok.. divide by x^4+x^2-80/x+3... not sure what to divide by here?

are you dividing (x^4+x^2-80)/(x+3)?.. if so

x+3 into x^4 +x^2-80... divide x into the x^4..

get x^3

.. multiply and subtract (x+3)*x^3

(x^3)(x+3) = x^4 +3x^3

x^4 + 0x^3 + x^2 -80
-(x^4 +3x^3)
----------------------------
0 - 3x^3 +x^2 -80

x +3 into -3x^3 + x^2 -80 ----------> divide next term by x and get

-3x^2 ------multiply by x+3 and subtract this from the remainder above
.......................... (x+3)(-3x^2) = -3x^3 - 9x^2

- 3x^3 +x^2 -80
-(-3x^3 - 9x^2)............. or change signs and add 3x^3 + 9x^2
--------------------
0 + 10x^2-80

now divide this next term by x and get 10x

multiply (x+3) by 10x and get = 10x^2 +30x.. subtract this from remainder above...

10x^2+0x-80
-(10x^2+30x)
----------------
0 - 30x - 80..................divide next term by x..... -30.. multiply by x+3 and subtract...

-30(x+3) = -30x -90

-30x - 80
-(-30x - 90)................. or change signs and add
---------------
0 +10................ this is the remainder.. it is not divisible by x

so.. your answer for this part.. is:

x^3 - 3x^2 + 10x - 30 R 10

Answer 7

You can set it up like a regular division problem. Note that I provided an x^3 term and an x term with zero coefficients because they will show up in the problem.
-------------------------------
x + 3) x^4 + 0x^3 + x^2 + 0x -80

Now divide x into x^4 and do the multiplication.

x^3
--------------------------------
x + 3) x^4 + 0x^3 + x^2 + 0x -80
x^4 + 3x^3

Like regular multiplication, subtract and bring down the next column.

x^3
--------------------------------
x + 3) x^4 + 0x^3 + x^2 + 0x -80
x^4 + 3x^3
---------------------
- 3x^3 + x^2

Keep on in this manner.

x^3 - 3x^2 + 10x - 30
---------------------------------
x + 3) x^4 + 0x^3 + x^2 + 0x -80
x^4 + 3x^3
---------------------
- 3x^3 + x^2
- 3x^3 - 9x^2
----------------------
10x^2 + 0x
10x^2 + 30x
--------------
- 30x - 80
- 30x - 90
------------
10

This comes out with a remainder of 10.

Perhaps the last term of the original dividend is really - 90?

(I just noticed that the system elinates leading blanks - I don't know how to get around that).

Answer 8

you are not allowed to devide to 0... it is NOT infinity - it is undefined (you may not say how many 0 are there in 5, if you add a lot of 0 one another you will never get 5)

the second one is 2 (everyone had solved it)

the second one was solved also by everyone = 25aa-30ab+9bb

and the division - when you devide you get xxx-3xx+10x-30 (and 10 left) so xxxx+xx-80=(x+3)(xxx-3xx+10x-30) + 10

Answer 9

x/0 is undefined
13+(-11) is positive 2, not negative
(5a-3b)² = 15a²-30ab+9b²
[x^4+x^2-80x^(-1)] ÷ (x+3). Use long division.

Answer 10

-5/0 is a number so large it can't be identified, therefore undefined. However, -5/0 can also be identified as negative infinity.