The rule would be ~y=4x+3~
Hope this helps
Have a great day/night
Feel free to ask any questions
The given equation has no solution when K is any real number and k>12
We have given that
3x^2−4x+k=0
△=b^2−4ac=k^2−4(3)(12)=k^2−144.
<h3>What is the condition for a solution?</h3>
If Δ=0, it has 1 real solution,
Δ<0 it has no real solution,
Δ>0 it has 2 real solutions.
We get,
Δ=k^2−144 here Δ is not zero.
It is either >0 or <0
Δ<0 it has no real solution,
Therefore the given equation has no solution when K is any real number.
To learn more about the solution visit:
brainly.com/question/1397278
F(x+1) = f(x) + 2.5 with f(x) is the current number.
Answer: A. A=(1000-2w)*w B. 250 feet
C. 125 000 square feet
Step-by-step explanation:
The area of rectangular is A=l*w (1)
From another hand the length of the fence is 2*w+l=1000 (2)
L is not multiplied by 2, because the opposite side of the l is the barn,- we don't need in fence on that side.
Express l from (2):
l=1000-2w
Substitude l in (1) by 1000-2w
A=(1000-2w)*w (3) ( Part A. is done !)
Part B.
To find the width w (Wmax) that corresponds to max of area A we have to dind the roots of equation (1000-2w)w=0 ( we get it from (3))
w1=0 1000-2*w2=0
w2=500
Wmax= (w1+w2)/2=(0+500)/2=250 feet
The width that maximize area A is Wmax=250 feet
Part C. Using (3) and the value of Wmax=250 we can write the following:
A(Wmax)=250*(1000-2*250)=250*500=125 000 square feets
