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3 years ago

Which angle is a vertical angle with ZEFD?

Mathematics

1 answer:

AVprozaik [17]3 years ago

6 0

Answer: ZAFB

Step-by-step explanation:

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Write the expression in algebraic form. [hint: sketch a right triangle.] tan(arcsec(x/3))

WARRIOR [948]

Sketch a right triangle having adjacent side(A) is given as “3”, hypotenuse side (H) is “x” and assigning angle “a” as the angle between A and H. Using Pythagorean theorem, you will get “square root of x-squared minus 9” as the opposite side (O). Using SOH CAH TOA function, and since secant is the reciprocal of cosine, sec(a) = x/3. Thus, a = arcsec(x/3). The remaining expression tan(a) is Opposite side over Adjacent side which is equal to “square root of x^2 - 9” over "3". Therefore, the algebraic expression would be: tan(arcsec(x/3)) = “sqrt (x^2 -9)” /3. Different answers can be made depending on which side you consider the “3” and “x”.

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3 years ago

Kona’s mother realized that her age is the difference of Kona’s age squared and four. If k represents Kona’s age, which expressi

san4es73 [151]

(K²+4)= her moms age :)

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3 years ago

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You are in charge of erecting a radio telescope on a newly discovered planet. To minimize interference, you want to place it whe

strojnjashka [21]

Answer:

Step-by-step explanation:

$\text{The equation of the surface of the sphere whose radius is 6 can be represented as:}$

$x^2+y^2+z^2 = 36}$

$So , g(x,y,z) = x^2 + y^2 + z^2 - 36$

$\text{To minimize the function :} \\ \\ M(x,y,z0 = 6x - y^2 +xz + 60$

$\text{by applying lagrange multipliers }$

$\bigtriangledown M (x,y,z) = \lambda \bigtriangledown g(x,y,z)$

$\langle 6+zz,-2y,x \rangle = \lambda \langle 2x,2y,2z \rangle$

$6+z = 2\lambda x$

$-2y = 2 \lambda y$

$\text{from the second equation ;} (\lambda +1)y = 0, \text{so , it is either y =0 or }\lambda = -1$

$Suppose \ \lambda = -1 ; \text{other equation becomes x = -2z and 6+z = -2x}$

$\text{such that; 6+z =4z or z = 2}$

$\text{it implies that: x= -4 and }y = \pm \sqrt{36 - 4-16} = \pm 4$

$\text{Here; there exist two possible points}$

$\text{(-4,-4,2) and (-4,4,2)}$

$\text{In the scenario here y=0,} \lambda \text{ is unknown, then we remove it}$

$x = 2 \lambda z \\ \\ 6+z = 2 \lambda x \\ \\ 6+z = 4 \lambda ^2z \\ \\ 6 = (4 \lambda ^2 -1) z ---(1)$

$z = \dfrac{6}{4 \lambda ^2 -1}$

$x = \dfrac{12 \lambda }{4 \lambda ^2 -1 }$

$\text{recall that; it i possible to divide }4 \lambda ^2 -1 \text{since (1) shows that it cannot be equal to zero.} \\ \\ \text{hence, puttinf this into constraint, whereby y =0, Then:}$