Answers to the fifth grade Chinese Evaluation Manual (Volume 2)

Answers to the fifth grade Chinese Evaluation Manual (Volume 2)

When the wind blows through the barren mountains and mountains, there are waves of forest, where there are birds

There are three square pools, large, medium and small. Their inner sides are 6 meters, 3 meters and 2 meters respectively. When two piles of gravel are sunk in the water of the medium and small pools, the water surface of the two pools rises by 6 cm and 4 cm respectively. If these two piles of gravel are sunk in the water of the large pool, how many cm does the water surface of the large pool rise?

6cm = 0.06m, 4cm = 0.04m. 3 × 3 × 0.06 = 0.54 (M3), 2 × 2 × 0.04 = 0.16 (M3), 0.54 + 0.16 = 0.7 (M3). The bottom area of the large pool is 6 × 6 = 36 (M2). The water surface of the large pool has increased by 0.7 △ 36 = 0.736 (m) = 3518 (CM). A: the water surface of the large pool has increased by 3518 cm

How to calculate the covariance?
Three variables X, y, Z. cov (Z, x) = 10, cov (Z, y) = 5
A = 2X + Y - 1
What is cov (Z, a) equal to?

From the covariance property cov (x1 + X2, y) = cov (x1, y) + cov (X2, y), we get
Cov(Z,A)=Cov(Z,2X+Y-1)=2Cov(Z,X)+Cov(Z,Y)-0=25
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Simple calculation question 1: 3 / 5 × 3.7 + 3.6 + 6.3 × 5 / 3 question 2: (3 / 4 × 7.2 + 2.8 × 0.75) △ 1.5
Question 3 6 / 11 (1 / 2 + 4 / 5 + 8 / 15)
Question 4: 117.4 - 17.4 △ 0.4 + 0 △ 7.61
When a fruit store brings back a batch of fruit, it sells 35% of the fruit on the first day, and 60 kg more on the second day than on the first day. At this time, there is still 1 / 4 of this batch of fruit left. How many kg of this batch of fruit was brought back? (write the equation, write the process)
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3/5×3.7+3.6+6.3×3/5
=3/5×（3.7+6.3）+3.6
=3/5×10+3.6
=6+3.6
=9.6
（3/4×7.2+2.8×0.75）÷1.5
=[3/4×（7.2+2.8）]×2/3
=3/4×2/3×10
=1/2×10
=5
6/11÷（1/2+4/5÷8/15）
=6/11÷（1/2+4/5×15/8）
=6/11÷（1/2+3/2）
=6/11÷2
=3/11
117.4-17.4÷0.4+0÷7.61
=117.4-43.5+0
=73.9
Set: this batch of fruit was originally transported back to us in a total of x kg
35%x+35%x+60+1/4x=x
(1-0.25-0.35-0.35)x=60
0.05x=60
x=1200

lim(1/√(n²+n)+1/√(n²+2n)+… +1 / √ n & # 178; + n & # 178;, n tends to infinity

The original formula = LIM (n - > ∞) [(1 / N) / √ (1 + 1 / N) + (1 / N) / √ (1 + 2 / N) +. + (1 / N) / √ (1 + n / N)] (numerator and denominator divide by n) = LIM (n - > ∞) {(1 / N) [1 / √ (1 + 1 / N) + 1 / √ (1 + 2 / N) +. + 1 / √ (1 + n / N)]} = ∫ DX / √ (1 + x) (Application of definite integral definition) = [2 √ (1 + x)] = 2 (√ 2-1)

Then (a + b) + CB=

B=-A
B=1/C
（A+B）+CB
=A-A+C*1/C
=1

It is known that 3sin β = sin (2 α + β), and it is proved that Tan (α + β) = 2tan α

It is proved that the condition is: 3sin [(α + β) - α] = sin [(α + β) + α], and it is expanded as follows: 3sin (α + β) cos α - 3cos (α + β) sin α = sin (α + β) cos α + cos (α + β) sin α, that is, 2Sin (α + β) cos α = 4cos (α + β) sin α

It is known that the cubic power of (- 3x to the fourth power, y to the third power) is divided by (- 1.5x to the nth power, y to the second power) = - MX to the eighth power, y to the seventh power,

Hello: (- 3x's fourth power, Y's third power) divided by (- 1.5x's nth power, Y's second power) = - MX's eighth power, Y's seventh power, 27x's 12th power, Y's 9th power △ 1.5x's nth power, Y's 2nd power = MX's 8th power, Y's 7th power, Y's 7th power = MX's 8th power, Y's 7th power, M = 1812-n = 8N = 12

In linear algebra, what is zero solution and what is non-zero solution

The zero solution is X1 = x2 = X3 = =xn=0
Non zero solution means that XM is not equal to 0
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The general formula of 5-3x + 2x square = 0

2x²-3x+5=0