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
Q2 Anwser
Exact Form:
28/15
Decimal Form:
1.86
Mixed Number Form:
1 13/15
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! :)
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.
