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dmitriy555 [2]
3 years ago
14

Are these triangles congruent?

Mathematics
1 answer:
Arada [10]3 years ago
6 0

Answer:

yes those triangles are congruent

Step-by-step explanation:

You might be interested in
Solve y = x - 5 if the domain is -3
DedPeter [7]

Answer:

y = -8

Step-by-step explanation:

The domain is the x values so x=-3

y = x-5

Substituting x=-3

y = -3-5

y=-8

6 0
3 years ago
A circle is translated 4 units to the right and then reflected over the x-axis. Complete the statement so that it will always be
irga5000 [103]

Answer:

The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

Step-by-step explanation:

Let C = (h,k) the coordinates of the center of the circle, which must be transformed into C'=(h', k') by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:

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

Where r is the radius of the circle, which remains unchanged in every operation.

Now we proceed to describe the series of operations:

1) <em>Center of the circle is translated 4 units to the right</em> (+x direction):

C''(x,y) = C(x, y) + U(x,y) (Eq. 1)

Where U(x,y) is the translation vector, dimensionless.

If we know that C(x, y) = (h,k) and U(x,y) = (4, 0), then:

C''(x,y) = (h,k)+(4,0)

C''(x,y) =(h+4,k)

2) <em>Reflection over the x-axis</em>:

C'(x,y) = O(x,y) - [C''(x,y)-O(x,y)] (Eq. 2)

Where O(x,y) is the reflection point, dimensionless.

If we know that O(x,y) = (h+4,0) and C''(x,y) =(h+4,k), the new point is:

C'(x,y) = (h+4,0)-[(h+4,k)-(h+4,0)]

C'(x,y) = (h+4, 0)-(0,k)

C'(x,y) = (h+4, -k)

And thus, h' = h+4 and k' = -k. The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

<em />

4 0
3 years ago
Factor out the gcf 63+45b
DerKrebs [107]

Answer:

GCF is 9

Step-by-step explanation:

Factor out 9 and divide what's in the parentheses by 9

9(7+5b)

3 0
3 years ago
Fraction simple 25/100
steposvetlana [31]

Answer:1/4

Step-by-step explanation:

i just thinked it

3 0
3 years ago
Distance between points 3,3 and 5,7?
ohaa [14]

Answer:  d = 4√5  or 4.47

Step-by-step explanation:

To find the distance of two-point, we can use the distance formula:

d=\sqrt{(x_{2} -x_{1} )^2 +(y_{2} -y_{1} )^2}\\\\d=\sqrt{(5 -3 )^2+ (7 -3)^2}\\\\d=\sqrt{(2)^2+ (4)^2}\\\\d=\sqrt{4+16}\\\\d=\sqrt{20}\\\\d = \sqrt{4*5}\\\\\\d = 4\sqrt{5}\\\\\\or d ≈ 4.47

7 0
3 years ago
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