(10x + 4) + (5x-4) ≈ 180
solve for x and you get x≈12
plug 12 in to 5x-4 and you get 56
Answer:
We know that the equation of the circle in standard form is equal to <em>(x-h)² + (y-k)² = r²</em> where (h,k) is the center of the circle and r is the radius of the circle.
We have x² + y² + 8x + 22y + 37 = 0, let's get to the standard form :
1 - We first group terms with the same variable :
(x²+8x) + (y²+22y) + 37 = 0
2 - We then move the constant to the opposite side of the equation (don't forget to change the sign !)
(x²+8x) + (y²+22y) = - 37
3 - Do you recall the quadratic identities ? (a+b)² = a² + 2ab + b². Now that's what we are trying to find. We call this process <u><em>"completing the square"</em></u>.
x²+8x = (x²+8x + 4²) - 4² = (x+4)² - 4²
y²+22y = (y²+22y+11²)-11² = (y+11)²-11²
4 - We plug the new values inside our equation :
(x+4)² - 4² + (y+22)² - 11² = -37
(x+4)² + (y+22)² = -37+4²+11²
(x+4)²+(y+22)² = 100
5 - We re-write in standard form :
(x-(-4)²)² + (y - (-22))² = 10²
And now it is easy to identify h and k, h = -4 and k = - 22 and the radius r equal 10. You can now complete the sentence :)
Root 25 is the rational number as it equals 5
Answer:
P(x< 18) = 0.986
Step-by-step explanation:
Step 1: find the z-score using the formula, z = (x - µ)/σ
Where,
x = randomly chosen values = 18
µ = mean = 7
σ = standard deviation = 5
Proportion of the population that is less than 18 = P(x < 18)
Plug in the values into z = (x - µ)/σ, to get z-score.
z = (18 - 7)/5
z = 11/5 = 2.2
Step 2: Find P(x< 18) = P (z<2.2) using z-table.
The probability that corresponds with z-score calculated is 0.986.
Therefore,
P(x< 18) = 0.986
Answer:
a) n<1 and n>5
b) 0 < n < -4
c) n > 2 and n < -2
Step-by-step explanation:
The signal is given by x[n] = 0 for n < -1 and n > 3
The problem asks us to determine the values of n for which it's guaranteed to be zero.
a) x[n-2]
We know that n -2 must be less than -1 or greater than 3.
Therefore we're going to write down our inequalities and solve for n
Therefore for n<1 and n>5 x [n-2] will be zero
b) x [n+ 3]
Similarly, n + 3 must be less than -1 or greater than 3
Therefore for n< -4 and n>0, in other words, for 0 < n < -4 x[n-2] will be zero
c)x [-n + 1]
Similarly, -n+1 must be less than -1 or greater than 3
Therefore, for n > 2 and n < -2 x[-n+1] will be zero
