16x^2-vox=0 when x=2 so
16x^2-vox=0
64-2vo
64=2vo
vo=32 ft/s
Answer:
(x, y) = (1/2, -1)
Step-by-step explanation:
Subtracting twice the first equation from the second gives ...
(2/x +1/y) -2(1/x -5/y) = (3) -2(7)
11/y = -11 . . . . simplify
y = -1 . . . . . . . multiply by y/-11
Using the second equation, we can find x:
2/x +1/-1 = 3
2/x = 4 . . . . . . . add 1
x = 1/2 . . . . . . . multiply by x/4
The solution is (x, y) = (1/2, -1).
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<em>Additional comment</em>
If you clear fractions by multiplying each equation by xy, the problem becomes one of solving simultaneous 2nd-degree equations. It is much easier to consider this a system of linear equations, where the variable is 1/x or 1/y. Solving for the values of those gives you the values of x and y.
A graph of the original equations gives you an extraneous solution of (x, y) = (0, 0) along with the real solution (x, y) = (0.5, -1).
Answer:

which agrees with the first answer in the list of possible options.
Step-by-step explanation:
We can use the fact that the addition of all four internal angles of a quadrilateral must render
. Then we can create the following equation and solve for the unknown "h":

Therefore the angles of this quadrilateral are:

Factor out a 3x:
3x(x2+4x+6)
Since you can't factor out the quadratic that's as far as it goes:
Answer:
3x(x2+4x+6)
Answer and Step-by-step explanation:
Considering the table attached.
(a) over 9.5 kg;
μ = 8
σ = 0.9
z = 9.5 - 8/0.9 ≈ 1.67
P (Z > 1.67) = 0.5 - P(0<Z<1.67) = 0.5 - 0.4525 = 0.0475
(b) at most 8.6 kg;
z = 8.6-8/0.9 ≈ 0.67
P(Z < 0.67) = 0.5 + P(0<Z<0.67) = 0.5 + 0.2486 = 0.7486
(c) between 7.3 and 9.1 kg.
z₁ = 7.3-8/0.9 ≈ -0.78
z₂ = 9.1 - 8/0.9 ≈ 1.22
P(-0.78 < Z < 1.22) = P(0 < Z < 0.78) + P(0 < Z < 1.22) = 0.2823 + 0.3888 = 0.6711