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alukav5142 [94]
3 years ago
11

Which of the following are congruence transformation? Check all that apply

Mathematics
2 answers:
Zinaida [17]3 years ago
8 0
The correct answer to this question is:

A-"Rotation"
D-"Reflecting"
E-"Translating"

Hope this helped, Mollifay450
Your Welcome :)
Black_prince [1.1K]3 years ago
5 0

Answer with explanation:

The Meaning of "Congruence transformation" is that the image after either  of any transformation, 1. Reflection,2. Rotation,3. Translation,4.and ,Dilation (Stretching and Shrinking) ,the geometrical shape should be congruent.

Now, Suppose, A triangle has gone through ,all the four transformation

→So, when you will reflect triangle through any of the line, preimage and image will be congruent, neither the shape ,nor size of the image changes.

→Similarly, when you will rotate or translate a geometrical shape,that is triangle, the Image and Preimage will be congruent.

→But, When we dilate ,a geometrical shape, if Dilation factor >1, Geometrical shape expands,that is get Stretched,and if,0< Dilation factor < 1, geometrical shape,shrinks.The two preimage and image will be Similar,not Congruent, when Dilated.

So, From the Options Provided, following are congruent transformation.

A- Rotation

D- Reflecting

E-translating

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Answer:

3 hours

Step-by-step explanation:

43-16=27

27/9=3

8 0
3 years ago
How does graphing linear inequalities differ from graphing linear equations?​
Ivan

Explanation:

When the inequality symbol is replaced by an equal sign, the resulting linear equation is the boundary of the solution space of the inequality. Whether that boundary is included in the solution region or not depends on the inequality symbol.

The boundary line is included if the symbol includes the "or equal to" condition (≤ or ≥). An included boundary line is graphed as a solid line.

When the inequality symbol does not include the "or equal to" condition (< or >), the boundary line is not included in the solution space, and it is graphed as a dashed line.

Once the boundary line is graphed, the half-plane that makes up the solution space is shaded. The shaded half-plane will be to the right or above the boundary line if the inequality can be structured to be of one of these forms:

  • x > ...   or   x ≥ ...   ⇒ shading is to the right of the boundary
  • y > ...   or   y ≥ ...   ⇒ shading is above the boundary

Otherwise, the shaded solution space will be below or to the left of the boundary line.

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Just as a system of linear equations may have no solution, so that may be the case for inequalities. If the boundary lines are parallel and the solution spaces do not overlap, then there is no solution.

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The attached graph shows an example of graphed inequalities. The solutions for this system are in the doubly-shaded area to the left of the point where the lines intersect. We have purposely shown both kinds of inequalities (one "or equal to" and one not) with shading both above and below the boundary lines.

6 0
3 years ago
In ΔPQR, m∠P = (3x+16) ,m∠Q=(4x-15),m∠R(2x+8) What is the value of x?
noname [10]

Answer:

  x = 19

Step-by-step explanation:

Assuming all angle measures are in degrees, the sum of them is 180:

  (3x +16) +(4x -15) +(2x +8) = 180

  9x +9 = 180 . . . . collect terms

  x + 1 = 20 . . . . . . divide by 9

  x = 19 . . . . . . . . . .subtract 1

The value of x is 19.

3 0
3 years ago
Solve by factoring 6p^2 =21p-6
uysha [10]

Given,

The expression is:

6p^2=21p-6

By using the middle term splitting method,

\begin{gathered} 6p^2-21p+6=0 \\ 2p^2-7p+2=0 \\ (x-(\frac{7+\sqrt{33}}{4}))(x+\frac{7+33}{4})=0 \end{gathered}

Hence, the factor is (x-(7+sqrt(33)/4)) (x+(7+sqrt(33)/4)).

7 0
1 year ago
You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta
vladimir1956 [14]

Answer:

The solutions listed from the smallest to the greatest are:

x:  -\sqrt{2}   -\sqrt{2}  \sqrt{2}  \sqrt{2}

y:      -1         1     -1     1

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The slope of the tangent line at a point of the curve is:

m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }

m = -\frac{\cos 2\theta}{\sin \theta}

The tangent line is horizontal when m = 0. Then:

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\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right), for all i \in \mathbb{N}_{O}

\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}, for all i \in \mathbb{N}_{O}

The first four solutions are:

x:   \sqrt{2}   -\sqrt{2}  -\sqrt{2}  \sqrt{2}

y:     1        -1        1     -1

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x:  -\sqrt{2}   -\sqrt{2}  \sqrt{2}  \sqrt{2}

y:      -1         1     -1     1

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