Hello here is a solution :
x²-8y-6x+y²=4....(1)
A)
x²-6x = x²-2(3)x+9-9
= x²-2(3)x+3²-9
x²-6x =(x-3)² -9
B)
y²-8y = y²-2(4)y +4²-16
y²-8y = (y-4)²-16
C)
in (1) :
(x-3)² -9 +(y-4)²-16 = 4
(x-3)² +(y-4)² = 29 ...<span>is the standard form of the equation
</span> <span>Select answer 1 :
</span><span>A: 3
B: 4
C: (x−3)2+(y−4)2=29</span>
If x < - 2 and x ∈ { - 5, - 2.6, - 2, - 0.8, 1, 1.5 }
- 5 < - 2 and - 2.6 < - 2
The solution set is:
A ) { - 5 , - 2.6 }
The answer to the question is D
Strange question, as normally we would not calculate the "area of the tire." A tire has a cross-sectional area, true, but we don't know the outside radius of the tire when it's mounted on the wheel.
We could certainly calculate the area of a circle with radius 8 inches; it's
A = πr^2, or (here) A = π (8 in)^2 = 64π in^2.
The circumference of the wheel (of radius 8 in) is C = 2π*r, or 16π in.
The numerical difference between 64π and 16π is 48π; this makes no sense because we cannot compare area (in^2) to length (in).
If possible, discuss this situatio with your teacher.
Let’s take fred age as x and Natalie’s age as y.
So,
A/Q
X + Y = 43. Let it be eqution 1
and
X + 4Y = 79. Let it be equation 2
Combining equation 1 and 2, we get
( X + Y) + 3Y = 79
43 + 3Y = 79
3Y = 36
Y = 12
Now, evaluating the value of Y in equation 1 we get,
X + 12 = 43
X = 31
Therefore, Fred is 31years old and Natalie is 12 years old.
