MATHEMATICS MIDDLE SCHOOL

Which pair of triangles can be proven congruent by SAS? 2 identical triangles are shown. The triangle is reflected across a line to form the second triangle. 2 identical triangles are shown. The triangle is rotated 90 degrees to form the second triangle. 2 identical triangles are shown. The triangle is reflected across a line and then rotated to form the second triangle. 2 identical triangles are shown. The triangle is rotated up and to the left 90 degrees to form the second triangle.

Answers

Answer 1
Answer:

Answer:

its the third option

Step-by-step explanation:

the one with 2 rings on the second longest angle 2 lines on longest side one side and 1 line on the shortest side

Answer 2
Answer:

Final answer:

All the described pairs of triangles that are reflected, rotated, or both, can be proven congruent by the SAS postulate because such rigid transformations preserve the congruency of sides and angles.

Explanation:

To determine which pair of triangles can be proven congruent by the SAS (Side-Angle-Side) postulate, we look for two sides of a triangle and the included angle that are congruent to two sides and the included angle of another triangle. When a triangle is reflected across a line, rotated 90 degrees, or both reflected and rotated, it maintains its size and shape, thus the corresponding sides and angles remain congruent.

A reflection or rotation (including a combination of both) is a type of rigid transformation which preserves the size and shape of figures. Hence, all the given pairs of triangles can be proven congruent to the original triangle through the SAS postulate, as rigid transformations do not alter the congruity of sides and included angles.

Learn more about SAS Postulate here:

brainly.com/question/2819482

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So we will multiply 13(1/2)6 = 39

Area = 39
The area is 39m, because using the formula for a triangle, 1/2bh, you can find this. :)

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Answers

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Answers

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FIRST ANSWER GETS BRAINLIESTRichard simplified an expression in three steps, as shown: x to the power of negative 4 multiplied by y to the power of 2, over y multiplied by x to the power of 2 multiplied by x to the power of 2 multiplied by y to the power of negative 4, the whole to the power of 3 equals x to the power of negative 12 multiplied by y to the power of 6, over y to the power of 3 multiplied by x to the power of 6 multiplied by x to the power of 6 multiplied by y to the power of negative 12 equals x to the power of negative 12 multiplied by y to the power of 6, over y to the power of negative 36 multiplied by x to the power of 36. Which is the first incorrect step and why?

Step 1, all the exponents are multiplied by 3

Step 2, the exponents in the denominator are multiplied

Step 3, the exponents of the same base are subtracted during division

Step 3, the exponents of the same base are added during division

Answers

Answer:

Step 3, the exponents of the same base are added during division

Step-by-step explanation:

Name 3 fractions that are equivalent to: 2/3Name 3 fractions equivalent to: 3/4

Name 3 fractions equivalent to:4/5

Answers

For 2/3 there are equvialent fractions such as 4/6,6/9,8/12 
for 3/4 there is 6/8,9/12,12/16
for 4/5 there is 8/10,12/15,16/20
These can all be figured out by multiplying the numerator and the denomanator by the same number! Each of these fractions can be reduced back to there original fraction! Hope this helps
Three fractions equivalent to ⅔ are 4/6 , 8/12, and 16/24
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What is the HCF of 2 x y and 6 x y?

Answers

(2 x y) is a factor of (6 x y), so it's the highest common factor of both expressions.
3y is the ans to your question
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