Agree with the girl making the bows because to make three bows all you need is 3/5 feet of ribbon, which isn’t even a full foot
Answer:
x=2
Step-by-step explanation:
let the number be x
9x-8=6+2x
9x-2x=6+8
7x=14
x=14/7
x=2
Y = 1/4x - 1....the slope here is 1/4....but for a perpendicular line we need a negative reciprocal slope...all that means is " flip " the slope and change the sign. 1/4....flip it....4/1.....change the sign....-4/1 or just -4. So our perpendicular line will have a slope of -4. The only one in ur answer choices with a slope of -4 is D.....so ur answer has to be D.
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.
A) I use x=-5; f(x)=-11 and x=-4; f(x)=-3. We get the range = -3-(-11)=8b)I use x=-5; f(x)=-11 and x=-3; f(x)=5. We get the range = 5-(-11)=16c)I use x=-5; f(x)=-11 and x=-2; f(x)=13. We get the range = 13-(-11)=24d) Range of input is equal with the ratios of the output. You can find the pattern of output above 8,16,and 24 can divided by 8 and give the result 1,2 and 3
Is that good for you?
Good Luck!
<3
