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3 years ago

Which theorem or postulate proves that △ABC and △DEF are similar?

Mathematics

2 answers:

GaryK [48]3 years ago

5 0

For similarity measures of corresponding angles should be equal so we would use AA similarity postulate

Svetlanka [38]3 years ago

4 0

Answer:

AA similarity postulate holds here.

Step-by-step explanation:

As in right angled ΔCAB and ΔFDE we are given two angles each and with the help of property that in a triangle the sum of all the angles of a triangle is equal to 180° w can find out the measure of third angle.

In ΔCAB we have:

∠A=90° and ∠B=37°.

The measure of third angle i.e. ∠C will be:

∠A+∠B+∠C=180°

90°+37°+∠C=180°

∠C=180°-127°

⇒ m∠C=53°

Similarly in ΔFDE we get:

m∠E=37° ( since we are given ∠F=53° and ∠D=90°)

Hence clearly all the corresponding angles in both the triangles are equal.

Hence AAA similarity holds.

Hence, both the triangles are similar.

Hence, first option is true.

AA similarity postulate

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babunello [35]

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We will apply the unitary method to solve this question

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irina [24]

Using how tall it is over how wide
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valentina_108 [34]

Answer:

E. This polynomial could be factored by using grouping or the perfect squares methods.

Step-by-step explanation:

x^2 + 2x + 1

There is no greatest common factor

This is a perfect square

a^2 + 2ab+ b^2 = ( x+1)^2

We can factor this by grouping

x^2 + 2x + 1

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aalyn [17]

Answer:

x = 1 or x = -1

Step-by-step explanation:

Given equation:

$x^{100}-4^x \cdot x^{98}-x^2+4^x=0$

Factor out -1:

$\implies -1(-x^{100}+4^x \cdot x^{98}+x^2-4^x)=0$

Divide both sides by -1:

$\implies -x^{100}+4^x \cdot x^{98}+x^2-4^x=0$

Rearrange the terms:

$\implies 4^x \cdot x^{98}-4^x-x^{100}+x^2=0$

$\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c \quad \textsf{to }x^{100}:$

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Factor the first two terms and the last two terms separately:

$\implies 4^x(x^{98}-1)-x^2(x^{98}-1)=0$

Factor out the common term $(x^{98}-1)$ :

$\implies (4^x-x^2)(x^{98}-1)=0$

Zero Product Property: If a ⋅ b = 0 then either a = 0 or b = 0 (or both).

Using the Zero Product Property, set each factor equal to zero and solve for x (if possible):

$\begin{aligned}x^{98}-1 & = 0 & \quad \textsf{or} \quad \quad4^x-x^2 & = 0 \\x^{98} & =1 & 4^x & = x^2 \\x & = 1, -1 & \textsf{no}& \textsf{ solutions for } x \in \mathbb{R}\end{aligned}$

Therefore, the solutions to the given equation are: x = 1 or x = -1

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