x^2-16x+48=0 How to solve it

x^2-16x+48=0 How to solve it

x²-16x+48=0
(x-4)(x-12)=0
The solution is x = 4 or x = 12

In these three formulas, the divisible formula is (), and the divisible formula is ()

Put it away
6 △ 12 = 0.5, 91 △ 13 = 7, 23 △ 7 = 3.2, in these three formulas, the divisible formula is (91 △ 13 = 7), the divisible formula is (6 △ 12 = 0.5, 91 △ 13 = 7,)
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Derivation formula: (x + y) (x2-xy + Y2) = X3 + Y3

Left = x3-x2y + XY2 + x2y-xy2 + Y3 = X3 + Y3 = right

What is the unit conversion of density?

The commonly used density units are g / cm ^ 3 and kg / m ^ 31. G / cm ^ 3 = 10 ^ 3 kg / m ^ 3. The conversion process is 1 g / cm ^ 3 = 1 * 10 ^ (- 3) kg / [10 ^ (- 6) m ^ 3] = 10 ^ 3 kg / m ^ 3. 1 kg / m ^ 3 = 10 ^ (- 3) g / cm ^ 3. The conversion process is 1 kg / m ^ 3 = 1 * 10 ^ 3 g / (...)

Simple calculation of recurrence equation: 3.4 × 9 + 99 × 0.6
Step by step

3.4×9+99×0.6
=3.4x(10-1)+(100-1)x0.6
=34-3.4+60-0.6
=94-4
=90

If the focus of the parabola y = 4x is f and the coordinates of the point m are (4,4), then there are several circles passing through the points F, m and tangent to the Quasilinear l of the parabola

Two

The volume of cuboid and cylinder is equal, the area of cuboid bottom is 2 / 5 of that of cylinder bottom, so the height of cuboid is ()

Let the area of the cuboid be: X and the height be: H
The height of the cylinder is: H
Then there is the equation x * H = 5 / 2 * x * H
The results are as follows
h/H=5/2

How to calculate 99 times 62 + 62 with simple operation

（99+1）×62=6200

Proving Lagrange mean value theorem

It is proved as follows: if the function f (x) is differentiable on (a, b) and continuous on [a, b], then there must be a ξ ∈ [a, b] such that F & # 39; (ξ) * (B-A) = f (b) - f (a) and f (x) is y, so the formula can be written as △ y = F & # 39; (x + θ Δ x) * △ X & nbsp; (0 & lt; θ & lt; 1) & nbsp; & nbsp

A steel cylinder with a volume of 0.5m3 contains oxygen with a density of 6kg / m3. If one third of the oxygen is used in an electric welding, the mass of residual oxygen in the cylinder is () kg, and the density of residual oxygen is () kg / m3. For example, how to calculate the density of oxygen in this problem

The mass of oxygen in the cylinder is
M = ρ oxygen * V = 3kg
The quality of oxygen left after use is
m’=m*（1-1/3)=2kg
The density of residual oxygen is
ρ oxygen '= m' / v = 4 kg / m3
So the mass of oxygen left after use is 2kg, and the density of remaining oxygen is 4kg / m3