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bixtya [17]
3 years ago
15

}}{a^{4}b}" alt="\frac{a^{3}b^{5}}{a^{4}b}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
aliina [53]3 years ago
3 0

Answer:

b^4 / a

Step-by-step explanation:

I have attached the explanation above. hopefully this will help

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U= -7

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Find the area of trapezoid ABCD by decomposing it into a rectangle and triangle.
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Answer:

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Step-by-step explanation:

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3 years ago
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Find, corrrect to the nearest degree, the three angles of the triangle with the given vertices. A(1,0), B(3,6), C(-1,4)
Karo-lina-s [1.5K]

Answer:

45°, 45°, 90°

Step-by-step explanation:

Find the vectors to represent each sides

AB =<3-1, 6-0>=<2,6>  

AC = < -1-1, 4 - 0 > = < -2, 4>

Magnitude of the vectors

AB = √(2²+6²) = 6.32

AC = √ ((-2)² + 4²) = 4.47

cosθ = vector of AB × vector AC / ( Product of the magnitude of AB and AC) = 2 × (-2) + (6×4)/ (6.32×4.47) = 20 / 28.2504

θ = arcos(20 / 28.2504 ) approx = 45°

Magnitude of the vectors

BA =<1-3,0-6>=<-2,-6>

BC =<-1-3,4-6>=<-4,-2>

Magnitude of the vectors equals

BA = √((-2)² + (-6)²) = 6.325

BC = √((-4)² + (-2)²) = 4.4721

cosθ = (-2×-4) + (-6 ×-2) / (6.325 × 4.4721) = 20 / 28.286

θ  = arcos (20 / 28.286 ) = 45°

Magnitude of the vectors

CB =<3--1, 6-4>=<4,2>

CA=<1--1,0-4>=<2,-4>

Magnitude of the vector =

CB = √(4² + 2²) = 4.4721

CA = √(2² + (-4)²) = 4.4721

cosθ = (4×2) + (2×-4) / (4.4721×4.4721) = 0

θ = arcos 0 = 90°

4 0
3 years ago
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algol13

Answer:

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