>>For what operation is the set of integers NOT closed??

Question

A. For what operation is the set of integers NOT closed??
1. subtraction
2. division
3. multiplication
3. addition

B. Which one of the following expressions is equal to (x+2)to the 2nd power?

choices:
1. x squared+ 6x +9
2. xsquared+6
3. xsquared+ 3x+9
4. xsquared+9

C. Linda paid $28 for a sweater that was on sale for 30% off the original price. What was the original price of the sweater?

1. $84
2. $40
3. $56
4. $72

D.If the angles of triangle are represented by x degree,
(3x + 20)degree, and 6xdegreesign, the triangle must be??

1. obtuse
2. isosceles
3. acute
4. right

E. Solve for x:
x+3 / 3x = x / 12


thank you soo much if you try to answer this..i'll apreciate it...really...

actually its a take home test...i dont know the answer because it's an honors thing..I'm not in honors..lol..
THANK YOU...

Answer 1

>>For what operation is the set of integers NOT closed??
A. For what operation is the set of integers NOT closed??
1. subtraction
2. division
3. multiplication
3. addition

Division as
integer + integer is always an integer
integer - integer is always an integer
integer x integer is always an integer

BUT integer / integer is not always an integer ... eg 7/3 is NOT an integer ← option 2.

B. Which one of the following expressions is equal to (x+2)to the 2nd power?

choices:
1. x squared+ 6x +9
2. xsquared+6
3. xsquared+ 3x+9
4. xsquared+9

Well (x + 2)² = (x + 2)(x + 2)
= x(x + 2) + 2(x + 2)
= x² + 2x + 2x + 4
= x² + 4x + 4 ← none of the options

But if you meant (x + 3)² you'd get:

(x + 3)² = (x + 3)(x + 3)
= x(x + 3) + 3(x + 3)
= x² + 3x + 3x + 9
= x² + 6x + 9 ← option 1.

C. Linda paid $28 for a sweater that was on sale for 30% off the original price. What was the original price of the sweater?

1. $84
2. $40
3. $56
4. $72

So if the price was reduced by 30% then Linda paid 70% of the original price

ie 70% of original price = $28

So original price = $28 /70%
=$28/0.7
= $40 ← option 2.



D. If the angles of triangle are represented by x°,
(3x + 20)°, and 6x°, the triangle must be??

1. obtuse
2. isosceles
3. acute
4. right

Angle sum of triangle is 180°

So x + 3x + 20 + 6x = 180

ie 10x + 20 = 180
So 10x = 160
So x = 16
Thus the angles are

16°, (3*16 + 10)° and 6*16°

ie 16°, 58° and 96° (Note that the sum of these angles is 180°)

Since it has one obtuse angle (96°) and all angles are different the triangle is:
an obtuse angled (and scalene) triangle. ← option 1.

E. Solve for x:

x+3 / 3x = x / 12

So x + 1/x = x/12

Multiply all by 12x (which is the LCD of the fractions present)

12x² + 12 = x²

ie 11x² + 12 = 0
so 11x² = -12
ie x² = -12/11

This has NO real solutions as x² ≥ 0 for all real values of x

ALTERNATELY:
(x+3) / 3x = x / 12 ← in case you actually mean this.

Cross multiply:
3x * x = 12*(x + 3)

So x² = 4(x + 3) On divivding both sides by 3)

So x² - 4x = 12

ie x² - 4x -12 = 0

ie (x - 6)(x + 2) = 0
So x = -2 or 6

Answer 2

1). The integers aren't closed under division: 3 divided
by 4 is not an integer, for example.

2. None of these are equal to (x+2)² = x² + 4x + 4.
If you meant (x+3)² then you get x² + 6x + 9.

3. You set this up as follows:
Original price - discount = sale price
So
x - 0.3x = 28
0.7x = 28
x = 28/0.7 = 40.
Check 30% of 40 = 12,
40 -12 = 28.

4. The sum of the angles must be 180 degrees.
So x + 3x+20 + 6x = 180
10x+ 20 = 180
10x = 160
x = 16
3x + 20 = 68
6x = 96.
Since an angle exceeds 90 degrees,
The triangle is obtuse.

5. I'm not sure whether you wanted to write the left-hand side of
this as x + 3/3x = x/12
or (x+3)/3x = x/12,
so I'll do it both ways.
First, possibility
x + 3/3x = x/12.
Then 3/3x = 1/x.
x + 1/x = x/12.
(x²+1)/x = x/12
x² = 12x²+12
11x² = -12
x = ±√12/11i,
so no real solutions.

In the second case,
just cross multiply
3x² = 12x+36.
We could do this by quadratic formula, but I thought
of a quicker way.
Let's add x² to each side. What does that do?
It makes the right-hand side a square!
So 4x^2 = x²+12x+36 = (x+6)².
Take both possible square roots to get
2x = x + 6
x = 6.

The other possibility is
-2x = x + 6,
-3x = 6
x = -2.

I know that's a strange way to do that last one,
but it works!

Answer 3

A. Division. Notice that 1/2 is not an integer (even though 1 and 2 are)

B. (x+2)^2 = (x+2)(x+2) = x^2 + 2x + 2x + 4 = x^2 + 4x + 4
None of the choices given are correct.
However, if you modify your problem to (x+3)^2, you get:
(x+3)^2 = x^2 + 3x + 3x + 9 = x^2 + 6x + 9
So your answer would be 1.
Notice that in both cases, I used FOIL.

C. Let p = original price
28 = p - 30%*p
28 = p - 0.30*p = p(1-0.30) = 0.70*p
p = 28/0.70 = 40
So your answer would be 2.

D. Sum of all angles in a triangle must add up to 180.

x + 3x + 20 + 6x = 180
10x + 20 = 180
10x = 160
x = 16

Then the three angles are x = 16, 3x + 20 = 68, 6x = 96.
Since at least one angle is greater than 90 degrees (96 > 90), we can say that the triangle is obtuse. The answer should be 1.

E. Try cross multiplication and then simplify.

(x+3)/(3x) = x/12
12(x+3) = 3x^2
0 = 3x^2 - 12x - 36
0 = 3(x^2-4x-12)
0 = (x-6)(x+2)
x = 6 or -2

Answer 4

Hey, this looks like homework! Well, the first one is instructive, so we can talk about that one....

Basically, the question is whether or not you get another integer when you add/subtract/multiply/divide two integers. Thinking about it, it should be clear that division is the exception - you can't divide two integers and expect to get another integer. In fact, division is not closed for real numbers either, since division by zero produces an undefined result.

* * * * *

By the way, for the second question, I think you meant (x+3), not (x+2). Do the expansion, and I think you'll get it.

Answer 5

A comment on what zanti said. Real numbers is an ordered field. As such, every element, EXCEPT ZERO, has a multiplicative inverse. In higher math subtraction and division disappear to be replaced by

additive inverse
and
multiplicative inverse

Answer 6

A. 2. division

B. none of the above

C. 2. 40

D. 3. obtuse

E. x=6 or x=-2

I hope I was helpful.

Answer 7

hhhmmmmmmmm....

Thats a hard one!


sorry! don't know!