<h3>
Answer: x > -8</h3>
Multiply both sides by 2 so that you go from
x/2 > -4
to
x > -8
The inequality sign will not flip since we are not multiplying both sides by a negative number.
If you are curious how to graph the solution, draw out a number line and plot an open unfilled circle at -8. Then shade to the right of the open circle. The open circle says "do not include this value as part of the solution set". The solution set is shaded representing all of the values larger than -8 (which make the original inequality true).
Answer:
Jenna has 6 and 4/5 cookies left.
Step-by-step explanation:
Jenna starts with 10 cookies. She eats 1, her brother eats 2, and her little sister eats 1/5. If you add up all the cookies they ate. that would be 3 1/5 cookies they ate. So, 6 and 4/5 cookies are left.
ps. give me brainliest pls
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.
Answer: 1.25 or 1¼
Step-by-step explanation:
6¼÷5=6.25÷5
1.25 or 1¼ or 125%
