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stepan [7]
3 years ago
14

Write the types of triangles drawn below

Mathematics
2 answers:
zhenek [66]3 years ago
4 0
A. Right triangle (it has a right angle)
b. An equilateral triangle and equiangular triangle (equal sides means equal angles)
c. Right triangle (right angle)
d. Obtuse triangle (an angle greater than 90 degrees)
e. Isosceles triangle (two equal sides)
f. Scalent triangle (three different side lengths)
g. Acute triangle (angles less than 90 degrees)
h. Isosceles triangle (two equal sides because two angles are equal)
i. Equilateral and equiangular (any triangle that is equiangular is equilateral)
podryga [215]3 years ago
4 0
Right triangle
b. An equilateral triangle and equiangular triangle
c. Right triangle
d. Obtuse triangle
e. Isosceles triangle
f. Scalent triangle
g. Acute triangle
h. Isosceles triangle l
i. Equilateral and equiangular


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A.) the balance is payed off during the grace period  

Step-by-step explanation:

Paying only the minimum payment would cause the remainder of the balance to draw interest charges, whether paid during the grace period or before the due date.

Paying off the balance during the next billing cycle will still result in interest charges for the current billing cycle.

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I Need Your Help Y'all,​
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1. X=3
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The legs of an isosceles triangle measure ( 2 x^4 + 2 x − 1 ) units each. The perimeter of the triangle is ( 5 x^4 − 2 x^3 + x −
nirvana33 [79]

Answer:

The base is x^4 -2x^3 -3x-1

Step-by-step explanation:

We know the perimeter of a triangle is the sum of the three sides

P = s1+s2+s3

We know the perimeter is  5 x^4 − 2 x^3 + x − 3

and two of the legs are 2 x^4 + 2 x − 1  since it is an isosceles triangle

P = 2s1 + s3

Subtract 2s1 from each side

P-2s1 =2s1 +s3 -2S1

P -2s1 =s3

Substituting what we know

5 x^4 − 2 x^3 + x − 3 - 2(2 x^4 + 2 x − 1) = s3

Distribute the -2

5 x^4 − 2 x^3 + x − 3 - 4 x^4 -4 x + 2 = s3

Combine like terms

5 x^4-4x^4 − 2 x^3 + x  -4 x -3+ 2 = s3

x^4 -2x^3 -3x-1 =s3

The base is x^4 -2x^3 -3x-1

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If m∠ABC = 72°, what is m∠DAB?<br><br> A) 72°<br> B) 108°<br> C) 144°<br> D) 216°
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∠ABC=∠CDA=72º
∠DAB=∠BCD=x

∠ABC+∠CDA+∠DAB+∠BCD=360º

Therefore:
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So, <span>∠DAB=∠BCD=x=108º

Answer: </span>∠DAB=108º  (B)
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