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3 years ago

PLEASE HELP ME I BEG YOU!?

Mathematics

2 answers:

nikdorinn [45]3 years ago

8 0

Distance formula : d = sqrt (x2 - x1)^2 + (y2 - y1)^2
(1,-2)(-4,3)
d = sqrt (-4 - 1)^2 + (3 - (-2)^2
d = sqrt (-5^2) + (5^2)
d = sqrt (25 + 25)
d = sqrt 50
d = 7.07

d = sqrt (x2 - x1)^2 + (y2 - y1)^2
(-5,3)(-5,9)
d = sqrt (-5-5)^2 + (9 - 3)^2
d = sqrt (-10^2) + (6^2)
d = sqrt (100 + 36)
d = sqrt 136
d = 11.66

diagonal is sqrt w^2 + h^2
diagonal = sqrt (9^2 + 5^2)
diagonal = sqrt (81 + 25)
diagonal = sqrt 106
diagonal = 10.295

sorry.. thats about all I can do.....geometry is not my best subject

weeeeeb [17]3 years ago

3 0

The answer to question 5 is D.5 and the answer to question 9 is (-2,-6)

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Answer:

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Step-by-step explanation:

The reference angle in radians can be found by the formula ...

ref angle = min(mod(θ, π), π -mod(θ, π))

Equivalently, it is ...

ref angle = min(ceiling(θ/π) -θ/π, θ/π -floor(θ/π))×π

<h3>Application</h3>

When we divide 11 radians by π, the result is about 3.501409. The fractional part of this quotient is more than 1/2, so the reference angle will be ...

ref angle = (1 -0.501409)π radians ≈ 1.56637 radians ≈ 89.746°

<em>Additional comment</em>

For calculations such as this, you need to use the most accurate value of pi available. The approximations 22/7 or 3.14 are not sufficiently accurate to give good results.

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3 years ago

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Which coordinate pair identifies a point in the third quadrant of the coordinate plane?

Fudgin [204]

Answer:

D, (-6,-2)

Step-by-step explanation:

the first quadrant of a coordinate plane has the x coordinate being a negative number, and the y coordinate being a positive number

the second quadrant of a coordinate plane has the x and y coordinates being a positive number

the third quadrant of a coordinate plane has the x and y coordinates being a negative number

the fourth quadrant of a coordinate plane has the x coordinate being a positive number, and the y coordinate being a negative number

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Find the area. Do you multiply all sides?

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Answer:

Step-by-step explanation:

I think you are to assume that this is a trapezoid and that the top right angle symbol is missing.

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b2 = 6.2

h = 3

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Prove<br><br> that BUGS is a trapezoid with coordinates B(-2,-1), U(0,3), G(3,2) and S(4,-3).

telo118 [61]

Answer:

Proved

Step-by-step explanation:

Given

$B =(-2,-1)$

$U = (0,3)$

$G = (3,2)$

$S = (4,-3)$

Required

Prove BUGS is a trapezoid

Given the coordinates, to prove a trapezoid; all we need to do is to check if one pair of sides is parallel.

<u></u>

<u>Taking BU and GS as a pair</u>

First, we calculate the slope using:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

For BU

$B =(-2,-1)$ --- $(x_1,y_1)$

$U = (0,3)$ --- $(x_2,y_2)$

So, we have:

$m = \frac{3 - -1}{0- -2}$

$m = \frac{4}{2}$

$m = 2$

For GS

$G = (3,2)$ --- $(x_1,y_1)$

$S = (4,-3)$ --- $(x_2,y_2)$

So, we have:

$m = \frac{-3-2}{4-3}$

$m = \frac{-5}{1}$

$m = -5$

<em>The slope of BU and GS are not the same; hence, they are not parallel.</em>

<u>Taking BS and GU as a pair</u>

Calculate the slope

For BS

$B =(-2,-1)$ --- $(x_1,y_1)$

$S = (4,-3)$ --- $(x_2,y_2)$

So, we have:

$m = \frac{-3 - -1}{4- -2}$

$m = \frac{-2}{6}$

$m = -\frac{1}{3}$

For GU

$G = (3,2)$ --- $(x_1,y_1)$

$U = (0,3)$ --- $(x_2,y_2)$

So, we have:

$m = \frac{3-2}{0-3}$

$m = \frac{1}{-3}$

$m = -\frac{1}{3}$

The slope of BS and GU are the same; hence, they are parallel.

<em>BUGS is a trapezoid because BS and GU have the same slope</em>

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