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3 years ago

Under the rectangular survey system, one section is equal to? 36 acres, 64, 360, 640?

Mathematics

1 answer:

daser333 [38]3 years ago

7 0

idk this a really hard problrm

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spayn [35]

Answer:

Step-by-step explanation:

side angle side

just look at the angle with the red line in the corner and for the side look at the two marks in the same spot you know its right because you only need to flip it for them to look the same

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3 years ago

Read 2 more answers

Gavin, Olivia and Michael are writing a report together. Gavin has written 2/5 of the report

Free_Kalibri [48]

Answer:

If there is 5 peices of a report and Gavin has written 2 out of 5 of those pieces, then there is only 3 pieces left. Meaning Oliver and Michael will either split 1.5 a nd 1.5 of the 3 pieces or one will take 2 and the other will take one.

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4 years ago

The vertices of AABC are A(-4,4), B(-2,2), and C(-5,2). If AABC is reflected across the line

Fantom [35]

(-5,2), it is C bc i said it is

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3 years ago

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:

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

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

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

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

$\text{by applying lagrange multipliers }$

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

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

$6+z = 2\lambda x$

$-2y = 2 \lambda y$

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

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

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

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

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

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

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

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

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

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

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