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Compare the 500 power of 3, 400 power of 4 and 300 power of 5 Step by step is required
500 power of 3 = 5 * 100 power of 3 = (5 power of 3) 100 power of 243
400th power of 4 = 4 * 100th power of 4 = (4th power of 4) 100th power of 256
300 power of 5 = 3 * 100 power of 5 = (3 power of 5) 100 power of 125
So the 500 power of 3, the 400 power of 4, and the 300 power of 5. The largest is the 400 power of 4, the smallest is the 300 power of 5, followed by the 500 power of 3,
Comparison size: 500 power of 2, 400 power of 3, 300 power of 4
If it is greater than 1, the 500 power of 2 is greater than that of 3. If it is less than 1, the 400 power of 3 is greater
2^500/3^400=lg2^500/lg3^400=(500*lg2)/(400*lg3)=(5*lg2)/(4*lg3)=lg2^5/lg3^4=lg32/lg81
One in four plus three in four minus five in seven plus two in seven equals zero
No, the answer is four out of seven
Solving three problems of equation with factorization in junior three mathematics
(1).4(2x-1)^2=x(1-2x）
(2).x(3x-1)^2=2x(1-3x)
(3).x^2+x-12=0
（9x-4）（2x-1)=0
x1=4/9,x2=0.5
（3x^2+X)(3X-1)=0
X1=1/3,X2=0,X3=-1/3
(X+4)(X-3)=0
X1=-4,X2=3
What is 31 ° 48 'equal to
31°48′
=31°+48/60°
=31.8°
4X + 25 * 3 = 21.9
4x+25*3=21.9
4x=21.9-75
4x=-53.1
x=-53.1/4
x=-13.275
1, 18.35 - (8.35 + 20 / 7)
2. 1.2: x = 3: two thirds
14.25x 9 / 5 + 1.8x 3 / 4 = 513 / 20 + 27 / 20 = 274, 6.3 ^ [(9 / 2-0.07x50) ^ 3 / 2] = 6.3 ^ [(2 / 9-7 / 2) ^ 2 / 3] = 6.3 ^ (- 59 / 18 × 3 / 2) = 6.3 ^ (- 59 / 12) = - 398 / 295 2
In the triangle ABC, a = 30 degrees, B = 12, the area of triangle ABC = 18, then the value of (Sina + SINB + sinc) / (a + B + C) is?
If the ABC area of triangle is 18, s = 1 / 2bcsina = 1 / 2 * c * 12 * 1 / 2 = 18, C = 6, cosa = from cosine theorem, Sina, SINB and sinc can be obtained
Write down the 1993 natural numbers from 1 to 1993 in turn, one of which is 123456789101113141516?
1+2+3+4+.+1993
=(1+1993)x1993÷2
=1994x1993÷2
=1987021
1+9+8+7+0+2+1=28
28 △ 9 = 3 and 1
So: 123456789101113141516. The remainder of 9 is 1
The radii of the two concentric circles are root 2 and root 3 respectively. AB and CD are the chords of the two circles respectively. When the area of the rectangle is the largest, what is the value of ad
To maximize the area of the rectangle, the short side of the rectangle is the length of any side of the circumscribed square of the small circle, that is, ab = CD = 2 times the root 2, so the quadratic power of ad = the quadratic power of root 3 - the quadratic power of half of AB = 1, so ad = 1
RELATED INFORMATIONS
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