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FromTheMoon [43]
3 years ago
11

Is ABC~LMN?If so, name which similarity postulate or theorem applies.

Mathematics
2 answers:
Hitman42 [59]3 years ago
7 0
The answer is variant A.
Because:
if the two angles of a triangle are congruent to two angles of the other triangle , this triangles are similar
Ray Of Light [21]3 years ago
7 0

Answer:

(A) Similar- AA

Step-by-step explanation:

Given: ΔABC is similar to ΔLMN

To find: Similarity postulate

Solution:

Two triangles are similar either by two angles or by one angle or two sides or by two angles and one side.

In this question it is given that two angles are equal, thus

In ΔABC and ΔLMN, we have

∠A=∠L=50° (Given)

∠B=∠M=75°(Given)

Thus, by AA similarity postulate,

ΔABC is similar to ΔLMN.

Hence, both the given triangles are similar by the option A postulate that is AA similarity postulate.

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3 years ago
3x^2(x^3-1)+2x^3(2x^2+2) <br> Whats the simplified answer?
ahrayia [7]

Answer:

\large\boxed{7x^5+4x^3-3x^2}

Step-by-step explanation:

3x^2(x^3-1)+2x^3(2x^2+2)\qquad\text{use the distributive property:}\ a(b+c)=ab+ac\\\\=(3x^2)(x^3)+(3x^2)(-1)+(2x^3)(2x^2)+(2x^3)(2)\qquad\text{use}\ a^na^m=a^{n+m}\\\\=3x^{2+3}-3x^2+4x^{3+2}+4x^3\\\\=3x^5-3x^2+4x^5+4x^3\qquad\text{combine like terms}\\\\=(3x^5+4x^5)-3x^2+4x^3\\\\=7x^5+4x^3-3x^2

4 0
3 years ago
Rhombus EFGH is shown. What is the measure of ∠HGJ?
MaRussiya [10]

Answer:

∠HGJ = 35°

Step-by-step explanation:

Although the question is somewhat incomplete, I have tried to reconstruct the question as much as possible to aid understanding (attached is the pictorial representation of the rhombus).

Firstly, it is important to note that rhombuses are parallelograms which have all sides equal and whose diagonals are perpendicular to each other. In addition to this, their diagonals bisect one other.

Mathematically, |EF| = |FG| = |GH| = |HE|

Another property which rhombuses have is that their opposite angles are equal.

Mathematically, ∠E = ∠G, ∠H = ∠F

The adjacent angles in a rhombus are supplementary. This means that the sum of angles closest to each other is 180°.

Mathematically, ∠E + ∠F = 180°, ∠F + ∠G = 180°, ∠G + ∠H = 180°,

∠H + ∠E = 180°

The question gave us ∠GHE and asked us to find ∠HGJ (denoted as θ in the picture attached). Note that ∠GHE means ∠H. That is, ∠GHE ⇒ ∠H

∠GHE or ∠H = 110°

From the properties of rhombuses (earlier stated), we know that adjacent angles in a rhombus are supplementary

Mathematically,

∠GHE + ∠FGH ⇒ 110° + ∠FGH = 180° ⇒ ∠FGH = 180° - 110°

∠FGH or ∠G = 70°

To calculate for ∠HGJ, we divide ∠FGH by 2, we have:

∠HGJ = ∠FGH ÷ 2 = 70° ÷ 2

∠HGJ = 35°

3 0
3 years ago
Please help!!!!
tatyana61 [14]

Answer:

Step-by-step explanation:

From the figure attached,

If the figures shown are congruent (Equal in shape and size), length of all the sides must be equal.

Therefore, sides sides CD and GH must be equal in measures.

Distance between the two points (x_1,y_1) and (x_2,y_2) is given by,

d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Length of the side CD having ends C(-4, 4) and D(0, 5) will be,

CD = \sqrt{(-4-0)^2+(4-5)^2}

CD = \sqrt{(16+1)}

     = \sqrt{17}

Length of GH having ends G(0, -4) and H(4, -5) will be,

GH = \sqrt{(0-4)^2+(-4+5)^2}

      = \sqrt{16+1}

      = \sqrt{17}

Therefore, m(CD) = m(GH)

Similarly, we can prove all corresponding sides of two figures equal in measure.

This shows ABCD ≅ EFGH.

3 0
3 years ago
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