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vlada-n [284]
2 years ago
10

Trapezoid GHJK was rotated 180° about the origin to determine the location of G'H'J'K', as shown on the graph. What are the coor

dinates of pre-image point H?
(2, 3)
(–2, 3)
(3, 2)
(3, –2)

Mathematics
2 answers:
Stolb23 [73]2 years ago
5 0

Answer:

(2,3)

Step-by-step explanation:

ahrayia [7]2 years ago
4 0

Answer:

when a point M (h, k) is rotated about the origin O through 180° in anticlockwise or clockwise direction, it takes the new position M' (-h, -k). The signs of the coordinates change .

In the given graph point H' is (-3,-2) .The pre image of this point will be point H .Point H will have coordinates which will have signs of both coordinates opposite to that  of point H'.

Coordinates of point H will be (3,2)

Step-by-step explanation:

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Answer:

StartFraction 1 Over 15 EndFraction left-parenthesis 10 h plus 60 right-parenthesis equals StartFraction 1 Over 10 Endfraction left-parenthesis 10 h right-parenthesis

Step-by-step explanation:

The desired equation equates 1/15 of the amount Lindsay earned with 1/10 the amount without the bonus:

StartFraction 1 Over 15 EndFraction left-parenthesis 10 h plus 60 right-parenthesis equals StartFraction 1 Over 10 Endfraction left-parenthesis 10 h right-parenthesis

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<em>Comment on the form of the answer</em>

If you don't like answers in this form, please do not post questions in this form. Ordinary math symbols are much appreciated.

6 0
2 years ago
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Which of the following is the expansion of (3c + d2)6?
vovangra [49]

Answer: Option A) 729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12} is the correct expansion.

Explanation:

on applying binomial theorem,  (a+b)^n=\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} a^{n-r} b^r

Here a=3c, b=d^2 and n=6,

Thus, (3c+d^2)^6=\sum_{r=0}^{6} \frac{6!}{r!(6-r)!} (3c)^{n-r} (d^2)^r

⇒ (3c+d^2)^6= \frac{6!}{(6-0)!0!} (3c)^{6-0}.(d^2)^0+\frac{6!}{(6-1)!1!} (3c)^{6-1}.(d^2)^1+\frac{6!}{(6-2)!2!} (3c)^{6-2}.(d^2)^2+\frac{6!}{(6-3)!3!} (3c)^{6-3}.(d^2)^3+\frac{6!}{(6-4)!4!} (3c)^{6-4}.(d^2)^4+\frac{6!}{(6-5)!5!} (3c)^{6-5}.(d^2)^5+\frac{6!}{(6-6)!6!} (3c)^{6-6}.(d^2)^6

⇒(3c+d^2)^6= \frac{6!}{(6-)!0!} (3c)^6.d^0+\frac{6!}{(5)!1!} (3c)^5.d^2+\frac{6!}{(4)!2!} (3c)^4.d^4+\frac{6!}{(6-3)!3!} (3c)^3.d^6+\frac{6!}{(2)!4!} (3c)^2.d^8+\frac{6!}{(1)!5!} (3c).d^{10}+\frac{6!}{(0)!6!} (3c)^0.d^{12}

⇒(3c+d^2)^6=(3c)^6.d^0+\frac{720}{120} (3c)^5.d^2+\frac{720}{48} (3c)^4.d^4+\frac{720}{36} (3c)^3.d^6+\frac{720}{48} (3c)^2.d^8+\frac{720}{120} (3c).d^{10}+.d^{12}

⇒(3c+d^2)^6=729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}

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Answer:

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So for example, you want to do a model of a house, and in the backyard of the house there are 4 trees, then in the model of the house you also need to put 4 trees in the backyard (indifferent of the scale of the model).

Then the number of boulders in the really fountain should be the same as the number of boulders in the scale model of the fountain.

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Answer:

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Explanation:

Benito's error was using the equal sign (=) instead of the congruency symbol (≅).

The congruency symbol (≅) means that the elements (segments, angles or figures in general) have the same measure, i.e. they have equal lengths for the segments or equal measure for the angles.

For instance, it is an error saying that the segment AB is equal to the segment BC because, as you clearly see in the picture, they are not same; they have the same length but they are joining different points, that makes them different in essence, although they have the same length. They would be equal only if they are the same figure.

In mathematics, you must not say that two different segments or two different angles are equal but they are congruent, which means that their lengths are equal. The use of equal is reserved for numbers and variables, not for figures like segment, points, angles, polygons.

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Answer: I need a picture off the pool and measurements.

Step-by-step explanation: I need this in order to figure out the problem and give you a helpful answer.

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