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Korolek [52]
3 years ago
6

Select the correct answer from each drop-down menu. ∆ABC has vertices at A(12, 8), B(4, 8), and C(4, 14). ∆XYZ has vertices at X

(6, 6), Y(4, 12), and Z(10, 14). ∆MNO has vertices at M(4, 16), N(4, 8), and O(-2, 8). ∆JKL has vertices at J(14, -2), K(12, 2), and L(20, 4).
_________blank1___________are congruent. A _______blank2______ is a single rigid transformation that maps the two congruent triangles.

Blank 1 options. A. Triangle Abc Triangle XYZ
B.Triangle ABC Triangle MNO
C.Triangle JKL Triangle ABC
D.Triangle MNO Triangle XYZ

Blank 2 options: A. Reflection
B. Dialtion
C.Rotation
D. Translation

Mathematics
1 answer:
lakkis [162]3 years ago
3 0

Answer: The correct option for blank 1 is option B and the correct option for blank 2 is option C.

Explanation:

It is given that ∆ABC has vertices at A(12, 8), B(4, 8), and C(4, 14). ∆XYZ has vertices at X(6, 6), Y(4, 12), and Z(10, 14). ∆MNO has vertices at M(4, 16), N(4, 8), and O(-2, 8). ∆JKL has vertices at J(14, -2), K(12, 2), and L(20, 4).

Plot these vertices on a coordinate plane and draw the triangles as shown below figure.

Two triangles are congruent if the both triangle have same corresponding sides and same angles.

From the figure it is noticed that

AB=MN=8

BC=NO=6

\angle B=\angle N

By SAS rule ∆ABC and ∆MNO are congruent.

So, the correct option for blank 1 is B.

From the It is clearly noticed the when we rotate the triangle ∆ABC 90^{\circ} counterclockwise along point B, we get ∆MNO.

So, the correct option for blank 2 is C.

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Aneli [31]

Answer:

subtract 2.4 from both sides,  simplify to -3x=16.8, divide by -3, simplify to x=-5.6.

Step-by-step explanation:

-3x + 2.4 = 19.2

 v     <u>-2.4</u>   <u>-2.4</u>

 v       0      16.8

    <u>-3x</u>=<u>16.8</u>

     -3    -3

       x=-5.6

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<h3>Noelle</h3>
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3 years ago
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Anestetic [448]

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5 0
3 years ago
Graph: Y – 3=1/2(x +2)
tatuchka [14]

Answer:

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3 years ago
\int (x+1)\sqrt(2x-1)dx
Nezavi [6.7K]

Answer:

\int (x+ 1) \sqrt{2x-1} dx =  \frac{1}{3}(x+1) (2x - 1)^{\frac{3}{2} } - \ \frac{1}{15}(2x-1)^{\frac{5}{2}} + C

Step-by-step explanation:

\int (x+1)\sqrt {(2x-1)} dx\\Integrate \ using \ integration \ by\ parts \\\\u = x + 1, v'= \sqrt{2x - 1}\\\\v'= \sqrt{2x - 1}\\\\integrate \ both \ sides \\\\\int v'= \int \sqrt{2x- 1}dx\\\\v = \int ( 2x - 1)^{\frac{1}{2} } \ dx\\\\v =  \frac{(2x - 1)^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}} \times \frac{1}{2}\\\\v= \frac{(2x - 1)^{\frac{3}{2}}}{\frac{3}{2}} \times \frac{1}{2}\\\\v = \frac{2 \times (2x - 1)^{\frac{3}{2}}}{3} \times \frac{1}{2}\\\\v = \frac{(2x - 1)^{\frac{3}{2}}}{3}

\int (x+1)\sqrt(2x-1)dx\\\\   = uv - \int v du                              

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