Finding the area of parallelogram with vector Given that the module of vector a is 4, the module of vector B is 3, and their angle is 30 degrees, find the area of the parallelogram with vectors a + 2B and a-3b as sides. Since the arrow can't be typed, please note that the letter in the title is vector The answer is 30, because I know how to calculate, but I can't,

Finding the area of parallelogram with vector Given that the module of vector a is 4, the module of vector B is 3, and their angle is 30 degrees, find the area of the parallelogram with vectors a + 2B and a-3b as sides. Since the arrow can't be typed, please note that the letter in the title is vector The answer is 30, because I know how to calculate, but I can't,

S = | (a + 2b) × (a-3b) | = | (a × a-3a × B + 2B × a-6b × B | [x is vector product]
∵a×a＝b×b＝0 a×b＝-b×a.|a×b|＝|a||b|sin＜a,b＞
∴S＝|5b×a|＝5 |b×a|＝5|b||a|sin＜b,a＞＝5*3*4*1/2＝30.

(-4a^6-20a^4b^2-12a^3b+6a^3b^2)/[-(-2a)^3]

(-4a^6-20a^4b^2-12a^3b+6a^3b^2)/[-(-2a)^3]
=(-4a^6-20a^4b^2-12a^3b+6a^3b^2)/(-8a^3)
=a^3/2+5ab^2/2+3b/2-3b^2/4

Given a ^ 2 + 4A + 1 = 0, find (1) a ^ 2 + 1 / A ^ 2 (2) (A-2) (1 / A + 2) (3) a ^ 4 + 1 / A ^ 4
Solve!

We know that a ^ 2 + 4A + 1 = 0,
a+1/a=-4;
Find (1) a ^ 2 + 1 / A ^ 2
=(a+1/a)²-2
=16-2
=14;
(2)(a-2)(1/a+2)
=1+2a-2/a-4
=-3+2(a-1/a)
=-3+2×±√[(a+1/a)²-4]
=-3±2×2√3
=-3 + 4 √ 3 or - 3-4 √ 3;
(3)a^4+1/a^4
=(a²+1/a²)²-2
=14²-2
=196-2
=194;
I'm very glad to answer your questions. Skyhunter 002 will answer your questions
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If a + 23 = B4 = C + 56, and 2a-b + 3C = 21

Let a + 23 = B4 = C + 56 = k, then a = 3K-2, B = 4K, C = 6k-5, so 2 (3K-2) - 4K + 3 (6k-5) = 21, the solution is k = 2, so a = 6-2 = 4, B = 8, C = 7, so a: B: C = 4:8:7

If (a + 2) / 3 = (c + 5) / 6 = B / 4 and 2a-b + 3C = 21, try to find the values of a, B and C

Let (a + 2) / 3 = (c + 5) / 6 = B / 4 = K
Then a = 3K-2, B = 4K, C = 6k-5
Then 2 (3K-2) - 4K + 3 (6k-5) = 21, k = 2
so a=4,b=8,c=7

Given a + 2 of 3 = C of 6 + 5 = B of 4, and 2A + B + 3C = 21, find the value of a, B, C
It's 2A + B + 3C = 21

a/3+2=c/6+5=b/4;
A / 3 + 2 = C / 6 + 5, so a / 3 = C / 6 + 3, a = C / 2 + 9
b=2c/3+20;
2A + B + 3C = 21 so c + 18 + 2C / 3 + 20 + 3C = 21 so C = - 51 / 14
Take it in and get a, B

If a + 23 = B4 = C + 56, and 2a-b + 3C = 21

Let a + 23 = B4 = C + 56 = k, then a = 3K-2, B = 4K, C = 6k-5, so 2 (3K-2) - 4K + 3 (6k-5) = 21, the solution is k = 2, so a = 6-2 = 4, B = 8, C = 7, so a: B: C = 4:8:7

Decomposition factor: A ^ 4 + A ^ 2-4a-3

a^4+a^2-4a-3
=(a⁴＋2a²＋1)－(a²＋4a＋4)
=(a²＋1)²－(a＋2)²
=(a²＋1＋a＋2)(a²＋1－a－2)
=(a²＋a＋3)(a²－a－1)

How much is sin a times cos a - 0.2 cos 4A

From Sina * cosa = = 0.2, we get: sin2a = - 0.4,
So: cos4a = 1-2 (sin2a) ^ 2 = 1-2 * 0.16 = 0.68

As shown in the figure, in △ ABC, the circle O with ab as the diameter intersects BC at point D, CE is perpendicular to AB, intersects o at point E and f respectively, intersects AB at point G, and connects be and de (1) to prove that ∠ bed = ∠ BCE (2) if the angle ACB = 45 °, ab = radical 5, CD = 2, the length of be and EF can be obtained

Er. The first question has been proved. I write the second one. I'm doing this. I'm ashamed. EF hasn't thought of the method of proof. I'll prove be first, please forgive me. And I'll make the process simpler. I really don't have the time. You have to write in detail when you do it, because the angle ACB = 45 degrees, so ad = DC = 2, so BD =