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lisabon 2012 [21]
2 years ago
7

What is 1/2 • 1 • 2 please help

Mathematics
1 answer:
BabaBlast [244]2 years ago
8 0

Answer:

1

Step-by-step explanation:

1/2x1x2

1

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Please help me thank you. <br><br> show all work.
seraphim [82]
The length of OM is 4/3 the length of OJ, and the length of ON is 4/3 the length of OK. Thus, two pairs of sides from the triangles are proportional to each other.

Also, angle O from one of the triangles is equal to angle O of the other triangle because they are the same angle.

Thus, the two triangles are similar by SAS (side-angle-side) similarity theorem. This theorem is quite similar to the SAS congruence theorem.

To make a similarity statement, we just have to match corresponding parts when naming the triangle.

Similarity statement: ΔOJK~ΔOMN
3 0
2 years ago
What is the area of the shaded triangle?
irinina [24]

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A = 1/2 × b × h

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7 0
3 years ago
7y+65.7=2 true or false(don't answer that I just don't want this taken down(unless you do actually have the answer) is anyone in
Sophie [7]

It's false and same lol

8 0
2 years ago
Check answer please
Cerrena [4.2K]
The fourth or the D) Option is correct.

To find the new induced matrix via a scalar quantified multiplication we have to multiply the scalar quantity with each element surrounded and provided in a composed (In this case) 3×3 or three times three matrix comprising 3 columns and 3 rows for each element which is having a valued numerical in each and every position.

Multiply the scalar quantity with each element with respect to its row and column positioning that is,

Row × Column. So;

(1 × 1) × 7, (2 × 1) × 7, (3 × 1) × 7, (1 × 2) × 7, (2 × 2) × 7, (3 × 2) × 7, (1 × 3) × 7, (2 × 3) × 7 and (3 × 3) × 7. This will provide the final answer, that is, the D) Option.

To interpret and make it more interesting in LaTeX form. Here is the solution with LaTeX induced matrix.

\mathcal{A = \begin{bmatrix}1 & 0 & 3 \\ 2 & -1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}

\mathbf{\therefore \quad 7A = 7 \times \begin{bmatrix}1 & 0 & 3 \\ 2 & - 1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}

\mathbf{\therefore \quad \begin{bmatrix}7 \times 1 & 7 \times 0 & 7 \times 3 \\ 7 \times 2 & 7 \times -1 & 7 \times 2 \\ 7 \times 0 & 7 \times 2 & 7 \times 1 \\ \end{bmatrix}}

\therefore \quad \begin{\bmatrix}7 & 14 & 0 \\ 0 & -7 & 14 \\ 21 & 14 & 7 \end{bmatrix}

Hope it helps.
5 0
3 years ago
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