Answer:
Step-by-step explanation:
From the given information:
Since three sides of the garden need to be enclosed because the fourth side is a building.
The area is xy and the perimeter will be x + 2y
Calculating the perimeter in terms of x; we have:
P = x+2y = 232 ----- (21)
y = (232 - x) / 2
y = 116 - (x/2)
Now; to calculate the area since the value of y in terms of x is known.
A = xy
A = x[116 - (x/2)]
A=116x - (x^2/2) ----- (2)
Taking the derivative of eqaution (2); we have:
A' = 116 - x
Relating it as it is equal to zero, thereby solving for x
; we have:
116 - x = 0
x = 116
From equation (1)
P = 116 +2y = 232
2y = 232 -116
2y = 116
y = 116/2
y = 58
Thus; the area A = xy can be calculated as:
A=xy
A=116(58)
A= 6728
Simplify both sides of the equation. (Distribute)
<span><span><span><span>(−7.8)</span>(x)</span>+<span><span>(−7.8)</span>(6.5)</span></span>=−25.74 </span> <span><span><span><span>−7.8x </span>+ </span>−50.7</span>= −25.74 -------------> </span>−7.8x − 50.7= −25.74
<span>Add 50.7 to both sides.
</span><span><span><span>−7.8x </span>− 50.7 </span>+ 50.7 </span>= <span>−25.74 + 50.7 ---------------------> </span><span>−7.8x </span>= <span>24.96
</span><span>Divide both sides by -7.8.
</span><span><span>−7.8x /</span>−7.8 </span>= <span>24.96/<span>−<span>7.8
</span></span></span><span>
X= </span><span>−3.2 :D</span><span>
</span>
Answer: (1,-1)
Step-by-step explanation:
Midpoint of BC=(6+4)/2, (3–1)/2. =(5,1)
Slope of BC is (3+1)/4–6)= 4/-2 = -2
Slope of perpendicular bisector of BC =+1/2
Eqn of perpendicular bisector is : Y-1 =1/2 (x-5)
Y=1/2 •(x-5) +1
Midpoint of AB. (6–2)/2, (3–1)/2 ={2,1)
Slope of AB is(3+1)/(-2–6) = 4/-8 =-1/2
Slope of perpendicular bisector = +2
Eqn of perpendicular bisector is Y-1. =2( X-2)
Y=2X-4+1 = 2X -3
Solving Y=(X-5)/2 +1
& Y=2X-3
2X-3 =(x-5/2)+1
2X-4 =(x-5/2)
4X-8 = x-5
3X =3
X=1
Y= 2×1–3= -1
Circumcentre is(1,-1)
The I and II are linear function, so those have the same domain and the same range always. So the answer is B
Answer:
1
Step-by-step explanation:
Given that, 1 to the power of 3
Generally, the rules for exponentiation states that if l is any real number and m is a positive integer, then lm can be shown as:
lm = l × l × … × l (m times)
Hence, 1 to the power of 3 can be written as 13, where the number 1 is called the base, and 3 is the power or exponent of the expression.
Therefore, 13 = 1 × 1 × 1 = 1
Hence, 1 to the power of 3 comes as 1.
