Fast solution of equation 8.8+4x=40 0.5x-13.2×5=6 2.8×4+5x=17.2 9x-x=48 0.06+7x=0.72×3 3×(x-7)=0.6

Fast solution of equation 8.8+4x=40 0.5x-13.2×5=6 2.8×4+5x=17.2 9x-x=48 0.06+7x=0.72×3 3×(x-7)=0.6


8.8+4x=40
x=(40-8.8)÷4
x=7.8
0.5x-13.2×5=6
x=(6+66)÷0.5
x=144
2.8×4+5x=17.2
x=(17.2-11.2)÷5
x=1.2
9x-x=48
x=48÷(9-1)
x=6
0.06+7x=0.72×3
x=(2.16-0.06)÷7
x=0.3
3×(x-7)=0.6
x=0.6÷3+7
x=7.2
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5 / 4-x = 9 / 20


X = 5 / 4-9 / 20 general score x = 25 / 20-9 / 20 x = 16 / 20 approximate score x = 4 / 5
Please accept if you are satisfied



Given a, B ∈ R, and a ≠ B, a + B = 2, then the size relation of 1, AB, A2 + B2 / 2 is?


ab



It is proved that the heights of the two waists of an isosceles triangle are equal


ZM: as shown in the figure, in △ BDC and △ CEB, ∵ ∠ DBC = ∠ ECB, ∵ BDC = ∠ CEB = 90 °, BC = BC, ≌ BDC ≌ CEB, CD = be



The derivative of function f (x) = lgx
Why is the derivative of function f (x) = lgx equal to f (x) '= 1 / x
(there must be a process of proof)
I can't prove it, so I'll ask you


First of all, let's say (LNX) '= 1 / X
Do it by definition
f'(x)=lim[f(x+△x)-f(x)]/△x △x→0
=[ln(x+△x)-lgx]/△x
=ln[(x+△x)/x]/△x
=ln(1+△x/x)/△x
When △ x → 0
Obviously, △ X / X → 0
Then ln (1 + △ X / x) and △ X / X are infinitesimals of equal order
Ln (1 + △ X / x) ~ △ X / X
Substitute:
ln(1+△x/x)/△x=(△x/x)/△x=1/x
That is, f '(x) = 1 / X
Get proof



A classroom is 8 meters long, 5.4 meters wide, 4 meters high, with an area of 20 square meters of doors and windows
A classroom is 8 meters long, 5.4 meters wide, 4 meters high, with an area of 20 square meters of doors and windows. It is necessary to paint the surrounding and ceiling of the classroom. If 250 grams of paint are used per square meter, how many kilograms of paint are needed?


Area around the classroom and ceiling: 8 * 4 * 2 + 5.4 * 4 * 2 + 8 * 5.4 = 150.4
Subtract the area of doors and windows by 20,
The obtained area is the area to be coated: 130.4 square meters
25 kg per square meter
130.4 * 0.25 = 32.6kg



If a and B are rational numbers, and satisfy 2 (a) - 2Ab + (b) + 4A + 4 = 0, find the value of B + a (b)


a^2-2ab+b^2+a^2+4a+4=0
(a-b)^2+(a+2)^2=0
So a = b = - 2, so - 16



For a triangle test field, it is necessary to divide the test field into four small plots with equal area, plant four different fine varieties, design more than three different division schemes, and give explanations


As shown in the figure:



The middle point m of the chord PQ in the circle O passes through the point m as two strings AB and CD, and the strings AD and BC intersect PQ at x and Y respectively,
Verification: m is the midpoint of XY





Square of M + 4 Mn + 3 N=


m²+4mn+3n²=(m+n)(m+3n)
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