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DiKsa [7]
3 years ago
5

Which theorem(s) can you use to prove that the two triangles are congruent?

Mathematics
1 answer:
kolezko [41]3 years ago
4 0

Answer:

To prove the congruence between two triangles there are several postulates, that can be applied depending on the congruent parts of the triangles.

The postulates are:

<u>SAS (Side-Angle-Side):</u> This postulate states that if two triangles have two corresponding sides congruent and one corresponding angle congruent, we can conclude that those triangles are completely congruent.

<u>ASA (Angle-Side-Angle):</u> This postulate states that if two triangles have two corresponding angles congruent and one corresponding side congruent, then those triangles are congruent.

<u>SSS (Side-Side-Side): </u>This postulate states that if two triangles have all three corresponding sides congruent, then triangles are congruent.

<u>AAS (Angle-Angle-Side):</u> If two pairs of corresponding angles are congruent and one pair of corresponding sides is congruent, then both triangle are congruent each other.

<u>HL (Hypothenuse-Leg):</u> this postulate is applied when we have right angles. It states that two right triangles are congruent if they have one pair of corresponding leg congruent, and the hypothenuses are also congruent. We could conclude that those right triangles are congruent.

From these postulate, we can demonstrate the congruence between two triangles. Then, from that congruence, we deduct the congruence between each corresponding parts of those triangles. For example, if we demonstrate the congruence by AAS, we deduct that the third angle is also congruent, and the other two sides are also congruent.

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It looks like you're given

<em>F'(x)</em> = 3<em>x</em>² + 7

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and you're asked to find <em>F(b)</em> for the values of <em>b</em> in the list {0, 0.1, 0.2, 0.5, 2.0}.

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On the other hand, if you're expected to <em>approximate</em> <em>F</em> at the given <em>b</em>, you can use the linear approximation to <em>F(x)</em> around <em>x</em> = 0, which is

<em>F(x)</em> ≈ <em>L(x)</em> = <em>F</em> (0) + <em>F'</em> (0) (<em>x</em> - 0) = 5 + 7<em>x</em>

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A <em>better</em> numerical method would be Euler's method. Given <em>F'(x)</em>, we iteratively use the linear approximation at successive points to get closer approximations to the actual values of <em>F(x)</em>.

Let <em>y(x)</em> = <em>F(x)</em>. Starting with <em>x</em>₀ = 0 and <em>y</em>₀ = <em>F(x</em>₀<em>)</em> = 5, we have

<em>x</em>₁ = <em>x</em>₀ + 0.1 = 0.1

<em>y</em>₁ = <em>y</em>₀ + <em>F'(x</em>₀<em>)</em> (<em>x</em>₁ - <em>x</em>₀) = 5 + 7 (0.1 - 0)   →   <em>F</em> (0.1) ≈ 5.7

<em>x</em>₂ = <em>x</em>₁ + 0.1 = 0.2

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