Question 10 on page 8 of new mathematics curriculum standard for Grade 6 Volume 2

Last year, today in this gate,
The peach blossom on the human face is red
People don't know where to go,
Peach blossom still smile spring breeze

Answers to mathematics unit test of grade 6 (Volume 2)

A word of advice, the answers to these exercise books are basically unanswered. Especially for your unknown exercise books, I suggest you search on the Internet or buy an answer book (some exercise books have an answer book)

In general, the ending of stressed closed syllable words is auxiliary + yuan + auxiliary, but some are not. Please help me to list the counter examples of common words in junior high school that are not like this (nouns, verbs, adjectives are best classified). The more the better

There are two kinds of open syllables: absolute open syllables and relative open syllables. Absolute open syllables: syllables ending with vowels. Relative open syllables: consonants - 1, vowels - consonants - syllables ending with E (except re)

English translation
Page 96 3A translation of the third grade English textbook of people's Education Press

Columbia: where I come from, we are very casual about the time. If you tell a friend that you go back to their home for dinner, it doesn't matter if you arrive on time. For us, it's very important to spend time with family and friends. We often visit friends' home by the way

In △ OAB, the vector OA = a, the vector ob = B, and OD is the height on the edge of ab. if ad = YAB, then the real number y =? (represented by a and b)

Because od is perpendicular to AB, the number of OD vectors multiplied by ab vector = 0
And because od vector = (1-y) times a vector + y times b vector
AB vector = B vector - a vector
That is [(1-y) a + Yb]. (B-A) = 0
It is reduced to y = [the square of the module of vector a minus the number product of vector a and b] / [the square of the module of vector B plus the square of the module of vector a minus twice the number product of vector AB]

Help to solve several trigonometric function problems (to process)
1. In △ ABC, sinbsinc = cos & sup2; a / 2
2. If cos & sup2; (a -- B) - cos & sup2; (a + b) = 1 / 2, (1 + Cos2 α) (1 + cos2b) = 1 / 3, find Tan α tanb
3. Simplify cos & sup2; a + cos & sup2; (a - π / 3) + cos & sup2; (a + π / 3)
4. Given that 0 ≤ a < B < R < 2 π, cosa + CoSb + cosr = 0, Sina + SINB + SINR = 0, find b-a
5. Given that a and B are acute angles, and 3sin & sup2; a + 2Sin & sup2; b = 1,3sin & sup2; a = 2Sin & sup2; B, find a + 2B
6. It is known that a, B, R are acute angles, Tana / 2 = Tan cubic R / 2, 2tana = tanr, proving that a, B, R form arithmetic sequence
7. It is known that sina / SINB = cos (a + b), where a and B are acute angles, and the maximum value of tanb is obtained

1.sinBsinC=cos²(A/2)
sinBsinC=(1+cosA)/2
2sinBsinC=1+cosA
2sinBsinC=1-cos(B+C)
2sinBsinC=1-cosBcosC+sinBsinC
1=cosBcosC+sinBsinC=cos(B-C)
That is to say, △ ABC is isosceles
2.cos²(a-b)-cos²(a+b)
=[cos(a-b)+cos(a+b)][cos(a-b)-cos(a+b)]
=2cosacosb*2sinasinb
=2sinacosa*2sinbcosb
=sin2asin2b
=2tana/(1+tan²a) * 2tanb/(1+tan²b)=1/2
∴8tanatanb=(1+tan²a)(1+tan²b)
(1+cos2a)(1+cos2b)
=[1+(1-tan²a)/(1+tan²a)][1+(1-tan²b)/(1+tan²b)]
=2/(1+tan²a) * 2/(1+tan²b)
=4/(1+tan²a)(1+tan²b)
=1/(2tanatanb)
=1/3
∴tanatanb=3/2
3.cos²a+cos²(a-π/3)+cos²(a+π/3)
=cos²a+[cos(a-π/3)+cos(a+π/3)]²-2cos(a-π/3)cos(a+π/3)
=cos²a+[2cosacos(π/3)]²-2[cosacos(π/3)+sinasin(π/3)][cosacos(π/3)-sinasin(π/3)]
=cos²a+4cos²acos²(π/3)-2[cos²acos²(π/3)-sin²asin²(π/3)]
=cos²a+2cos²acos²(π/3)+2sin²asin²(π/3)
=3cos²a/2+3sin²a/2
=3(cos²a+sin²a)/2
=3/2
4.sina+sinb+sinr=0
sinr=-(sina+sinb)
cosa+cosb+cosr=0
cosr=-(cosa+cosb)
sin²r+cos²r=(sina+sinb)^2+(cosa+cosb)^2
=2+2(sinasinb+cosacosb)
=2+2cos(b-a)
=1
cos(b-a)=-1/2
∵0≤a＜b＜r＜2π,
∴b-a=2π/3
5. Wrong title, 3sin2a = 2sin2b
3sin²a+2sin²b=1
3sin²a=1-2sin²b
3sin²a=cos2b
3sin2a=2sin2b
sin2b=3sin2a/2
sin2b=3sinacosa
cos(a+2b)
=cosacos2b-sinasin2b
=cosa*(3sin²a)-sina*3sinacosa
=0
∵0

Abed and AFCD are parallelograms. AF and de intersect at right angles. The intersection is point G, Ag is 3, DG is 4. The area of parallelograms is 36. Find the perimeter of quadrilateral ABCD
It's quadrilateral abed and AFCD

AG = 3, DG = 4, Ag is the height of parallelogram abed, DG is the height of parallelogram AFCD, and the area of two parallelograms is 36, so de = AB = 12, CD = AF = 9
And AGD is a right triangle, so ad = be = CF = 5
When the extension line of CD and Ba intersects h, ch = 12, BH = 16 and BC = 20 can be obtained
The circumference of ABCD is ab + BC + CD + Da = 12 + 20 + 9 + 5 = 46

What's the English word for getting on the bus?

get on the bus

Given the area of acute triangle ABC is 8, ab = 4, AC = 5, find the length of BC

Method 1: s (△ ABC) = (1 / 2) ab × acsina = 8, ｜ (1 / 2) × 4 × 5sina = 8, ｜ Sina = 4 / 5. ∵ △ ABC is an acute triangle, ｜ cosa = √ [1 - (Sina) ^ 2] = √ (1-16 / 25) = 3 / 5

Question 10 on page 8 of new mathematics curriculum standard for Grade 6 Volume 2