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aalyn [17]
3 years ago
13

Which transformation rule maps ​ triangle ABC ​ onto triangle DEF ?

Mathematics
1 answer:
Darina [25.2K]3 years ago
4 0

The x-coordinate of D is 7 greater than that of A.

The y-coordinate of D is 3 less than that of A.

The appropriate choice for the description of the mapping is

... (x+7, y−3)

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Define least common multiple​
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Answer:

the smallest common multiple of two or more numbers

Step-by-step explanation:

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How do I use substitution to solve a system of linear equations?
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Answer:

the substitution method is one way to solve linear equation. the substitution method is when substitute the one y value with the other:

example :

x+y=10

y=10-x ( value of y)

2x+y=12 ( substitute the value of y in the equation )

2x+(10-x)=12

2x+10-x=12 ( now solve for x)

x=12-10

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x+y=10

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3 years ago
How to tell if a line is parallel or perpendicular by the equation?
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It'll be the opposite reciprocal in front of the X like instead of 3 it will be -1/3 so flip it and change the sign
 
5 0
4 years ago
Read 2 more answers
Write the equation of a line perpendicular to y= -5/6x-3 and whose y-intercept is (0,11).
worty [1.4K]

Answer:

y=6/5x+11

Step-by-step explanation:

Perpendicular lines have the opposite sign and reciprocal of the original line. Since the y-intercept is 11 we know that it is 6/5x+11

8 0
3 years ago
A circle has the equation 2x²+12x+2y²−16y−150=0.
KonstantinChe [14]

Answer: B. The coordinates of the center are (-3,4), and the length of the radius is 10 units.

Step-by-step explanation:

The equation of a circle in the center-radius form is:

(x-h)^{2} +(y-k)^{2}=r^{2} (1)

Where (h,k) are the coordinates of the center and r is the radius.

Now, we are given the equation of this circle as follows:

2x^{2}+12x+2y^{2}-16y-150=0 (2)

And we have to write it in the format of equation (1). So, let's begin by applying common factor 2 in the left side of the equation:

2(x^{2}+6x+y^{2}-8y-75)=0 (3)

Rearranging the equation:

x^{2}+6x+y^{2}-8y=75 (4)

(x^{2}+6x)+(y^{2}-8y)=75 (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of (a\pm b)^{2}=a^{2}\pm+2ab+b^{2}:

<u>For the first parenthesis:</u>

x^{2}+6x+b^{2}

We can rewrite this as:

x^{2}+2(3)x+b^{2}

Hence in this case b=3 and b^{2}=9:

x^{2}+2(3)x+3^{2}=x^{2}+6x+9=(x+3)^{2}

<u>For the second parenthesis:</u>

y^{2}-8y+b^{2}

We can rewrite this as:

y^{2}-2(4)y+b^{2}

Hence in this case b=-3 and b^{2}=9:

y^{2}-2(4)y+4^{2}=y^{2}-8y+16=(y-4)^{2}

Then, equation (5) is rewritten as follows:

(x^{2}+6x+9)+(y^{2}-8y+16)=75+9+16 (6)

<u>Note we are adding 9 and 16 in both sides of the equation in order to keep the equality.</u>

Rearranging:

(x-3)^{2}+(y-4)^{2}=100 (7)

At this point we have the circle equation in the center radius form (x-h)^{2} +(y-k)^{2}=r^{2}

Hence:

h=-3

k=4

r=\sqrt{100}=10

8 0
3 years ago
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