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weqwewe [10]
4 years ago
13

What is the answer to 12-2x6+2/2

Mathematics
2 answers:
Marysya12 [62]4 years ago
6 0
12-2*6+ \frac{2}{2}  \\  \\ =12-12+1 \\  \\ =1
Paha777 [63]4 years ago
5 0
12-2*6+2/2=12-(2*6)+(2/2) = 12-12+1 = 1
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Evalute the expression. 2(2+4)^0+3(18+6)^0<br> A) 0<br> B) 1<br> C) 2<br> D) 5
Anastaziya [24]

Answer:

D, 5

Step-by-step explanation:

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4 years ago
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Can someone help me with this problem? It’s Special Right Triangles: Decimal Answer. Round to the nearest tenth. Thank you ! 10
Triss [41]

Answer:

h = 1.4

c = 2.8

Step-by-step explanation:

For each problem, remember the special triangle side ratios then use a proportion. To solve, isolate the variable.

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Since two of the angles are 45, this is an isosceles triangle. All isosceles triangles have two equal sides that are not the hypotenuse.

In a right isosceles triangle, the ratio for regular side to hypotenuse is 1 to √2.

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For the triangle with the variable c:

The is an equilateral triangle cut in half because the angles are 30 and 60.

The side ratio of altitude to hypotenuse is √3 to 2.

\frac{\sqrt{3} }{2} =\frac{c}{4} \\\sqrt{3} = \frac{c}{2}\\2\sqrt3 = c

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5 0
4 years ago
Adam tried to compute the average of his 6 test scores. he mistakenly divided the correct sum of all of his test scores by 5, wh
kipiarov [429]
Sum/5 = 90
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3 years ago
Part 3000 of me forgetting math please help
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Step-by-step explanation:

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2 years ago
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Use a matrix to find the coordinates of the endpoints or vertices of the image of each figure under the given reflection.
hjlf
The vertices of the original quadrilateral can be written in matrix form using the vertex matrix. The vertex matrix is
     \begin{pmatrix}-5&-1&-3&-7\\ 4&-1&-6&-3\end{pmatrix}

To find the coordinates of the endpoints or vertices of the image of the given coordinate points reflected about the y-axis, we just need to multiply the transformation matrix by the vertex matrix. The transformation matrix for this particular problem is 
     \begin{pmatrix}-1&0\\ 0&1\end{pmatrix}

Multiplying the two matrices, we have
     \begin{pmatrix}-1&0\\ 0&1\end{pmatrix}\begin{pmatrix}-5&-1&-3&-7\\ 4&-1&-6&-3\end{pmatrix}=\begin{pmatrix}5&1&3&7\\ 4&-1&-6&-3\end{pmatrix}

Therefore, the coordinates of the endpoints or vertices of the image are (5,4), (1,-1), (3, -6) and (7, -3).
     
7 0
3 years ago
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