A container that can hold 1kg of water can hold 1kg of kerosene and sulfuric acid

A container that can hold 1kg of water can hold 1kg of kerosene and sulfuric acid

The mass is equal to the density multiplied by the volume, because the mass is certain, so the volume and density are inversely proportional. The density of kerosene is less than that of water, so it can't be loaded. The density of sulfuric acid is greater than that of water, so it can be loaded!

A container that can hold only 1 kg of water can hold______ Kg kerosene; a bottle that can hold 1kg alcohol can hold water______ (ρ kerosene = ρ alcohol = 0.8kg/l)

① ∵ρ = MV ∵ the volume of 1kg water is v = m ρ = 1kg 1000kg / m3 = 1 × 10-3m3, so the mass of kerosene it can hold is m = ρ v = 0.8 × 103kg / m3 × 1 × 10-3m3 = 0.8kg. ② the volume of container is v = m alcohol, ρ alcohol = 1kg 0.8 × 103kg / m3 = 1.25 × 10-3m3; m water = ρ water, v = 1 × 103kg / m3 × 1.25 × 10-3m3 = 1.25Kg

Five sides are two fifths of the square cardboard into a rectangle, the perimeter of the rectangle is (), the area is ()
kuai

14/5；4/5

When m = m, MX ^ 2 + 4x + 3 is a complete square

When MX ^ 2 + 4x + 3 = 0 has only one root, MX ^ 2 + 4x + 3 is a complete square
Δ=16-4*3m=16-12m=0,m=4/3

It is known that F1F2 is the two focal points of the ellipse x ^ 2 / 9 + y ^ 2 / 4 = 1, and P is on the ellipse. If △ pf1f2 is a right triangle, find the PZ coordinate of the point

When Pf1 ⊥ and F1F2, then p (- √ 5,0)
When PF2 ⊥ F1F2, then p (√ 5,0)
When Pf1 ⊥ PF2, that is, ∠ f1pf2 = 90
Let P (x, y), x ^ 2 / 9 + y ^ 2 / 4 = 1
According to the straight line vertical: Y / (x - √ 5) * y / (x + √ 5) = - 1 ②
① (2) → x, y

Any prime number greater than 6 divided by 6 must have a remainder, which will only be () and ()
Any prime number greater than 6 divided by 6 must have a remainder, which will only be () and ()
What's in brackets?

The remainder will only be (1) and (5)

Given the function f (x) = x + 1 / X (x > 0), we prove that if 0

Let x1

If the function f (x) = LG [x (x-3 / 2) + 1], X ∈ [1,3 / 2], find the maximum value of F (x)

The answer should be 0
f(x)=lg[x(x-3/2)+1]=lg(x²-3/2x+1)
The axis of symmetry of X & sup2; - 3 / 2x + 1 is equal to - B / 2A = -- 1.5 / 2 = 3 / 4
So x & sup2; - 3 / 2x + 1 is an increasing function on [1,3 / 2]
And because lgx is an increasing function
So when x = 3 / 2 is the maximum, the maximum of F (x) is 0

Only proper prime numbers () × () = 91, () + () + () = 15, () × () = 38 can be filled in brackets

（7）×（13）=91,
（3）+（5）+（7）=15,
（2）×（19）=38

Let the probability density of two-dimensional random variables (x, y) be f (x, y) = CX ^ 2Y, x ^ 2 & lt; Y & lt; 1, and the others be 0;
I want to ask why the integral region of FY (y) is equal to - √ y to √ y when we find the edge density function

Only calculate the area with non-zero probability density, as shown in the figure. The economic mathematics team will help you solve it. Please evaluate it in time