A cuboid glass jar is 40cm in length, 20cm in width and 30cm in height. The water depth is 18cm. Now, an iron ball is completely immersed in the water, and the water surface rises to 25cm to calculate the volume of the iron ball

A cuboid glass jar is 40cm in length, 20cm in width and 30cm in height. The water depth is 18cm. Now, an iron ball is completely immersed in the water, and the water surface rises to 25cm to calculate the volume of the iron ball


Volume = 40 x 20x (25-18)
=800x7
=5600 square centimeters
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A cuboid shaped glass container, 40cm in length, 30cm in width and 50cm in height, is filled with water
How many cm is the water surface from the mouth of the container in a cuboid container with a height of 45 cm?


45-(40×30×50)÷(60×40)
=45-60000÷2400
=45-25
=20 cm



There is a cuboid glass container measuring 40 cm in length and 35 cm in width. It is known that there is 36 cm deep water in the container. Now we need to put the container in another direction. This is the depth of the container


Does the east-west orientation change into the north-south orientation? Or 36 ah ~ 2333



If two curves y = x2 + ax + B and 2Y = - 1 + XY3 are tangent to points (1, - 1), then the values of a and B are ()
A. 0,2B. 1,-3C. -1,1D. -1,-1


Y '= 2x + A, y' | x = 1 = 2 + A for y = x2 + ax + B, y '| x = 1 = 2 + A for 2Y = - 1 + XY3, y' = Y32 − 3xy2 so y '| x = 1 = - 12 − 3 = 1, so there is 2 + a = 1, a = - 1 substituting point (1, - 1) coordinates into y = x2 + ax + B, there is - 1 = 1 + A + B, and a = - 1, so B = - 2 + 1 = - 1, so a = - 1, B = - 1, B = - 1, so D is selected



If the solution set of inequality X & # 178; + (k-1) x + 4 〉 0 is r, the value range of real number k is obtained


The solution set is r, which means that the image of quadratic function is all above the x-axis, so b-square-4ac
That is, (k-1) 2-16



How to calculate this question vertically? 24 + 35=
It's about calculation --


24
+35
-----
fifty-nine
Add the addend bit 4 and the addend bit 5 to get 9,
Add the addend ten bits 2 and the addend ten bits 3 to get 5



Given LIM (x → 1) g (x) = LIM (x → 1) H (x) = 2, and G (x) ≤ f (x) ≤ H (x), then LIM (x → 1) [2x ^ 2 + 3f (x)] =?


From the condition g (x) ≤ f (x) ≤ H (x)
Because it is continuous, this inequality holds for any point of three functions
Then when X - > 1, we can know from the pinch theorem
lim(x→1)f(x)=2
be
lim(x->1) [2x^2+3f(x)]=2+6=8



6S = 27-5t 3S + 4T = 18 the following equations are solved by the method of addition, subtraction and elimination


6s+5t=27
6s+8t=36
t=3,s=2



X in sequence 1,1,2,3,5, x, 13,... Is equal to


x=3+5=8



Index and logarithm
Xiao Ming pointed out that equation 2 ^ x = x ^ 2 has two positive real number solutions (2,4) and one negative real number solution according to the image change trend of exponential function y = 2 ^ X and quadratic function y = x ^ 2. Xiao Hui said that both sides of the equation take the logarithm with 2 as the base, and they are transformed into 2log2 ^ x = x or log2 ^ x = 1 / 2 ^ X. according to the image, we can discuss how many real number solutions x ^ 2 = 2 ^ X has. Is Xiao Hui's method correct?


Xiaohui's method is wrong. First of all, the transformation is wrong. The logarithm with 2 as the base on both sides should be x = log 2 (base) x ^ 2, and then continue to transform to x = 2 log 2 (base) | x | the logarithm should be the absolute value of X, because the logarithm itself should be positive, the original x ^ 2 is positive, but after the index 2 is advanced, it should still be positive