Step-by-step explanation:
log2x = 1 - log(x+4)
log2x = log10 - log(x+4)
log2x = log[10/(x+4)]
2x = 10/(x+4)
2x(x+4) = 10
2x² + 8x - 10 = 0
÷2 both sides of equation
x² + 4x - 5 = 0
(x+5)(x-1) = 0
x = -5, 1
x isn't be -5 because logA, A need to be > 0
So, x value is 1
Given data:
The first set of equations are x+y=4, and x=6.
The second set of equations are 3x-y=12 and y=-6.
The point of intersection of first set of te equations is,
6+y=4
y=-2
The first point is (6, -2).
The point of intersection of second set of te equations is,
3x-(-6)=12
3x+6=12
3x=6
x=2
The second point is (2, -6).
The equation of the line passing through (6, -2) and (2, -6) is,
Thus, the required equation of the line is y=x-8.
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
Nineteen million, two hundred sixty-six thousand, four hundred twenty! :)
