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siniylev [52]
2 years ago
8

Find the area of the parallelogram with vertices A(−1,2,3), B(0,4,6), C(1,1,2), and D(2,3,5).

Mathematics
1 answer:
cupoosta [38]2 years ago
6 0

Answer:

5*sqrt3

Step-by-step explanation:

The vector AB= (0-(-1), 4-2,6-3) AB= (1,2,3)

The modul of AB is sqrt(1^2+2^2+3^2)= sqrt14

The vector AC is (1-(-1), 1-2, 2-3)= (2,-1,-1)

The modul of B is sqrt (2^2+(-1)^2+(-1)^2)= sqrt6

AB*AC= modul AB*modul AC*cosA

cosA=( 1*2+2*(-1)+3*(-1))/ sqrt14*sqrt6= -3/sqrt84=

sinB= sqrt (1- (-3/sqrt84)^2)= sqrt75/84= sqrt 25/28= 5/sqrt28

s= modul AB*modul AC*sinA= sqrt14*sqrt6* 5/ sqrt28= 5*sqrt3

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Answer:

(a) k'(0) = f'(0)g(0) + f(0)g'(0)

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Step-by-step explanation:

(a) Since k(x) is a function of two functions f(x) and g(x) [ k(x)=f(x)g(x) ], so for differentiating k(x) we need to use <u>product rule</u>,i.e., \frac{\mathrm{d} [f(x)\times g(x)]}{\mathrm{d} x}=\frac{\mathrm{d} f(x)}{\mathrm{d} x}\times g(x) + f(x)\times\frac{\mathrm{d} g(x)}{\mathrm{d} x}

this will give <em>k'(x)=f'(x)g(x) + f(x)g'(x)</em>

on substituting the value x=0, we will get the value of k'(0)

{for expressing the value in terms of numbers first we need to know the value of f(0), g(0), f'(0) and g'(0) in terms of numbers}{If f(0)=0 and g(0)=0, and f'(0) and g'(0) exists then k'(0)=0}

(b) m(x) is a function of two functions f(x) and g(x) [ m(x)=\frac{1}{2}\times\frac{f(x)}{g(x)} ]. Since m(x) has a function g(x) in the denominator so we need to use <u>division rule</u> to differentiate m(x). Division rule is as follows : \frac{\mathrm{d} \frac{f(x)}{g(x)}}{\mathrm{d} x}=\frac{\frac{\mathrm{d} f(x)}{\mathrm{d} x}\times g(x) + f(x)\times\frac{\mathrm{d} g(x)}{\mathrm{d} x}}{g^{2}(x)}

this will give <em>m'(x) = \frac{1}{2}\times\frac{f'(x)g(x) - f(x)g'(x)}{g^{2}(x) }</em>

on substituting the value x=5, we will get the value of m'(5).

{for expressing the value in terms of numbers first we need to know the value of f(5), g(5), f'(5) and g'(5) in terms of numbers}

{NOTE : in m(x), g(x) ≠ 0 for all x in domain to make m(x) defined and even m'(x) }

{ NOTE : \frac{\mathrm{d} f(x)}{\mathrm{d} x}=f'(x) }

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Answer:

Triangle APB is an isosceles triangle ⇒ 3rd answer

Step-by-step explanation:

* Lets explain the how to solve the problem

- ABCD is a square

∴ AB = BC = CD = AD

∴ m∠A = m∠∠B = m∠C = m∠D = 90°

- DPC is equilateral triangle

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