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Jobisdone [24]
4 years ago
14

Notebooks cost $1.20 each. This weekend they will be at a sale on for $.80, what percentage is the sale??

Mathematics
1 answer:
AlladinOne [14]4 years ago
7 0
0.8/1.2 = 0.6667
1 - 0.6667 = 0.3333

It is 33.33 % off
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What is a "real number"
grigory [225]

Answer: are numbers which  include all the rational numbers, such as -3 which is an integer and the fractions such as 5/4, and all the irrational numbers, such as √7, and π

6 0
3 years ago
This isn’t too difficult of questions and I am pretty sure I know the answers but I just want to make sure. Can someone please h
Shkiper50 [21]
Q2 Anwser
Exact Form:
28/15
Decimal Form:
1.86
Mixed Number Form:
1 13/15
5 0
3 years ago
Find the 10th term of the sequence defined by the given rule. Assume that the domain of each
velikii [3]

Answer: -10180

hope this helps

5 0
4 years ago
Abby filled her goodie bags with 4 cookies and 3 candy bars and spent a total of $10.25 per bag. Marissa filled her goodie bags
Paladinen [302]
Let's start by writing a system of linear equations:
c -> cookies 
cb -> candy bars
(You can use any abbreviations to your preference)

Abby:
4 cookies 
3 candy bars
$10.25 per bag
The equation would be:
4c+ 3cb = $10.25

Marissa:
2 cookies
7 candy bars
$14.75 per bag
The equation would be:
2c + 7cb = $14.75

So our linear equation system would be:
<span>4c+ 3cb = $10.25
</span><span>2c + 7cb = $14.75

I would try to get rid of one variable so I can solve for the other variable. In this case, it is easier to get rid of c since I can multiply the second equations by 2. Then it would subtract the two equations.

(2c + 7cb = $14.75) 2 = 4c + 14 cb = $29.50

      4c + 3cb = $10.25
   -  4c+14 cb = $29.50                    (4c would get canceled.)
---------------------------------
             -11 cb = - $19.25                (Divide by -11 to solve for cb)
</span>             ---------   -------------
              -11           -11
            
             cb = $1.75

Now we know cb (candy bar) cost, we would substitute this value into cb into one of the equations. It doesn't matter which equation you put it in. I will substitute it in the first equations. 

  4c + 3 (1.75) = $10.25
  4c + 5.25 = $10.25                       (Multiply 3 by 1.75)
       -5.25       -5.25                         (Subtract 5.25 on both sides)
             4c = 5                                 (Divide by 4 on both sides to get c)
             ----   ---
              4      4
         
         c= 1.25

Check the work:
4(1.25) + 3(1.75) 
  = $10.25

2(1.25) + 7(1.75)
   = $14.75

Total cost:
cookies = $1.25
candy bars = $ 1.75

Hope this helps! :)

6 0
4 years ago
Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k​
masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

2

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

2

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

2

−2k−4k+12

D=k^2-6k+13D=k

2

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0
3 years ago
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