Divide both sides by their LCD, 2
2/4 = 1/2
1/2 is irreducible
Answer:
Ix - 950°C I ≤ 250°C
Step-by-step explanation:
We are told that the temperature may vary from 700 degrees Celsius to 1200 degrees Celsius.
And that this temperature is x.
This means that the minimum value of x is 700°C while maximum of x is 1200 °C
Let's find the average of the two temperature limits given:
x_avg = (700 + 1200)/2 =
x_avg = 1900/2
x_avg = 950 °C
Now let's find the distance between the average and either maximum or minimum.
d_avg = (1200 - 700)/2
d_avg = 500/2
d_avg = 250°C.
Now absolute value equation will be in the form of;
Ix - x_avgI ≤ d_avg
Thus;
Ix - 950°C I ≤ 250°C
Answer:
40
Step-by-step explanation:
(2x+1/(2x))^5 *(2x -1/(2x))^5
= ((2x)^2 -1/(2x)^2)^5 (a+b)*(a-b) =a2-b2
= (4x^2-1/4(x)^2)^5
now
x =4x^2. ,a = 1/4(x)^2 ,n =5
we have
general term = Cr *x^r *a^(n-r)
= Cr * (4x^2)^r * (1/4(x)^2)^(n-r)
= Cr *4^r * X^2r * 1/( 4^(n-r) *x^(2n-2r)
= Cr * 4^r/4^(n-r) * x^(2r)/x^(2n-2r)
= Cr * 4(2r-n) *x(4r-2n)
now for x^2
4r-2n = 2
4r -10=2
4r =12
r = 3
now for coeff
C(5,3) * 4^(2*3-5)
5!/(3!*(5-3)!) * 4
5*4/(2*1)*4
40
So it work like this
180-72= 108
x+4=108
x=108-4
x=104
