Given that vectors a and B satisfy the following conditions: the module of vector a = 5, the module of vector b = 3, and the module of vector A-B is 7, find the point multiplication of vector a and B

Given that vectors a and B satisfy the following conditions: the module of vector a = 5, the module of vector b = 3, and the module of vector A-B is 7, find the point multiplication of vector a and B

Vector a, vector B and vector A-B can form a triangle with side lengths of 5, 3 and 7 respectively. Then the angle θ of vector a and B can be obtained by cosine theorem
cosθ=（|a|^2+|b|^2-|a-b|^2）/（2*|a||b|）
=-1/2
a.b=|a|*|b|*cosθ=-15/2

In the triangle ABC, the angle c is 90 degrees, and the central perpendicular BC of AB is at point D. if the angle bad angle DAC is 22.5 degrees, the degree of angle B is: A: 37.5 ° B: 67.5 ° C: 37.5 or 6

Let me answer: the key is to see an important condition: a triangle bad is an isosceles triangle
Then let the angle CAD be x, then the angle bad = angle DBA = x + 22.5
Then we get the formula 2x (x + 22.5) + X + 90 = 180 and x = 15
Angle B = 15 + 22.5 = 37.5
I don't care for the next one

Given that the function f (x) is a monotone increasing function on R and is an odd function, the sequence {an} is an arithmetic sequence, A3 ＞ 0, then the value of F (A1) + F (A3) + F (A5) ()
A. Constant positive B. constant negative C. constant 0d. Can be positive or negative

∵ function f (x) is an odd function on R and an increasing function sequence. Take any x2 > x1, there is always f (x2) > F (x1), ∵ function f (x) is an odd function on R, ∵ function f (0) = 0, ∵ function f (x) is an odd function on R and an increasing function, ∵ when x > 0, f (0) > 0, when x < 0, f (0) < 0. ∵ sequence {an} is an arithmetic sequence, a1 + A5 = 2A3, A3 > 0, ∵ a1 + A5 > 0 F (A1) + F (A5) > 0, ∵ f (A3) > 0, ∵ f (A1) + F (A3) + F (A5) are always positive numbers

On the topic of vector, judge right and wrong
1. The length of vector AB is equal to that of vector ba
2. If vector a is parallel to vector B, the directions of a and B are the same or opposite
3. Two vectors with the same starting point and the same end point must be the same
4. Two vectors with the same end point must be collinear
5. If vector AB and vector CD are collinear, then a.b.c.d must be on the same straight line
6. A directed line segment is a vector, and a vector is a directed line segment
Can you explain the cause of the mistake

1 pair
2 no, consider the zero vector
3 pairs
No, collinear vectors must be on the same line or parallel to each other
No, parallel is OK
Six is wrong, zero vector

(1 / 2) in the RT triangle ABC, the angle ACB = 90 degrees, AC = BC, D is the midpoint of BC, CE is perpendicular to AD and the extension line of BF / / AC intersection CE is perpendicular to F

〈ACD=〈CBF=90°,CE⊥AD,〈ECD+〈CDE=90°,〈CAD+〈CDA=90°,〈CAD=〈BCF,△CAD≌△BCF,BF=CD,CD=BD,BF=BD,〈DBG=〈CAB=〈FBG=45°,BG=BG,△DGB≌△FGB,〈FGB=〈DGB,〈DGB+〈FGB=180°,∴DF⊥AB...

Requirement: just write the answer directly, △ this is division sign × this is multiplication sign + - semicolon/
1: The speed of the train is 120 km / h, and the speed of the swallow is 150 km / h. what's the speed of the train? 2: 200 fruit trees were planted in the orchard this year, 198 of them survived. What's the survival rate of these fruit trees? 3: 196 trees survived in the hope primary school in spring, but 4 did not survive. What's the survival rate? 4: last year, 65 hectares of soybeans were planted in Yongfeng farm, 52 hectares of corn are planted. How much less is the area of corn planted than that of soybean planted? How much less is the area of soybean planted than that of corn planted? 5: the chemical fertilizer plant actually produced 50000 tons of chemical fertilizer in June, which is 5000 tons higher than that of the plan, and how much is the increase? (except for the endless percentages, reserve one before)

1 120÷150*100%=80%
2 198÷200=99%
3 196÷200=98%
4. The area of corn is less than that of Soybean (65-52) △ 65 = 20%
The area of soybean is more than that of corn (65-52) △ 52 = 25%
5 0.5÷（5-0.5）=11.1%

It is known that the area of triangle is 2, and there is p q in the plane of triangle ABC, which satisfies the vector PA + vector PC = 0
Vector QA = vector 2BQ, then the area of AQP is?

Vector PA + vector PC = 0
P is the midpoint of AC
Vector QA = vector 2BQ
Q is the trisection of Ba
Connect BP
P is the midpoint of AC
∴S△ABP=S△CBP=S△ABC*1/2=1
∵BQ=1/3AB
∴S△BPQ=1/3*S△ABP=1/3
∴S△APQ=1-1/3=2/3

As shown in the figure, ∠ BAC = 40 ° and ∠ B = 75 ° in △ ABC, ad is the angular bisector of △ ABC, and the degree of ∠ ADB is calculated

∵ ad bisection ∠ cab, ∠ BAC = 40 °, ∵ DAB = 12 ∠ BAC = 20 °, ∵ B = 75 °, ∵ ADB = 180 ° - ∠ DAB - ∠ B = 180 ° - 20 ° - 75 ° = 85 °

Two mathematical problems about fraction equation in grade two of junior high school,
1. If a / X-2 + 3 = 1-x / 2-x has an increasing root
So what is the value of a_____
2. We know that XY / x + y = 1 / 3, the square of X, the square of Y / x, the square of y = 1 / 5
Find the value of XY
Change the second question
Given XY / (x + y) = 1 / 3, x ^ y ^ / (x ^ + y ^) = 1 / 5, find the value of XY
^Denotes square
Question 2

1. To solve the equation, multiply both sides (X-2) to get a + 3 (X-2) = X-1, to get x = 5-a / 2, because the original equation has an increasing root, so X-2 = 0, to get x = 2, so 5-a / 2 = 2, to get a = 1 2. Wrong? Is x ^ 2Y ^ 2 / (x ^ 2 + y ^ 2) XY / (x + y) = 1 / 5, x + y = 5xy, x ^ + y ^ = (x + y) ^ - 2XY = 25X ^ y ^ - 2XY, so 1 / 3x ^ y ^ /

Let F 1 and F 2 be the left and right focal points of the ellipse x ^ 2 / 9 + y ^ 2 / 4 = 1 respectively. If P is on the ellipse and | vector Pf1 + PF2 | = 2 √ 5
Then the angle between vector Pf1 and vector PF2 is?

c^2=a^2-b^2=5,c=√5
F1(-√5,0),F2(√5,0)
Let P (m, n), then Pf1 = (- 5-m, - n), PF2 = (- 5-m, - n)
|PF1+PF2|=|（-2m,-2n）|=2|(m,n)|=2√(m^2+n^2)=2√5
That is m ^ 2 + n ^ 2 = 5
And P is on the ellipse, where m ^ 2 / 9 + n ^ 2 / 4 = 1
The solution: m ^ 2 = 9 / 5, n ^ 2 = 16 / 5
m=±3/√5,n=±4/√5.
Take P (3 / √ 5,4 / √ 5)
So Pf1 · PF2 = (- √ 5-3 / √ 5, - 4 / √ 5) (√ 5-3 / √ 5, - 4 / √ 5) = 9 / 5-5 + 16 / 5 = 0
So Pf1 ⊥ PF2