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PSYCHO15rus [73]
2 years ago
9

The proof TriangleABC ≅ TriangleDCB that is shown. Given: AngleA ≅ AngleD; CD||AB Prove: TriangleABC ≅ TriangleDCB Triangles C D

B and C A B are shown. Angles C D B and C A B are congruent. Sides D C and A B are parallel. What is the missing reason in the proof? A 2-column table has 5 rows. Column 1 is labeled Statement with entries angle A is-congruent-to angle D, line segment C D is parallel to line segment A B, line segment C B is-congruent-to line segment B C, angle A B C is-congruent-to angle D C B, triangle A B C is-congruent-to triangle D C B. Column 2 is labeled Reason with entries given, given, reflective property, alternating interior angles are congruent, question mark. alt. ext. Angles are ≅ ASA AAS corr. int. Angles are ≅
Mathematics
2 answers:
IrinaVladis [17]2 years ago
7 0

Answer:

AAS

Step-by-step explanation:

On edgu

Aloiza [94]2 years ago
5 0

Answer: AAS

Step-by-step explanation:

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Adiya’s method is not correct. To form a perfect square trinomial, the constant must be isolated on one side of the equation. Also, the coefficient of the term with an exponent of 1 on the variable is used to find the constant in the perfect square trinomial. Adiya should first get the 20x term on the same side of the equation as x2. Then she would divide 20 by 2, square it, and add 100 to both sides. 
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Answer:

75.

Step-by-step explanation:

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Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).
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Given : A florist currently makes a profit of $20 on each of her celebration bouquets and sells an average of 30 bouquets every week . and graph

To Find :  Maximum profit , breakeven point , profit interval

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peak of y from Graph

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at breakeven

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Monique made several batches of soup for a potluck supper. Each batch required Three-fourths of a pound of potatoes, and she use
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castortr0y [4]

For this case we have that a quadratic equation is of the form:

ax ^ 2 + bx + c = 0

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x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}

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4x ^ 2-3x + 9 = 2x + 1\\4x ^ 2-3x-2x + 9-1 = 0\\4x ^ 2-5x + 8 = 0

We look for the roots:

x = \frac {- (- 5) \pm \sqrt {(- 5) ^ 2-4 (4) (8)}} {2 (4)}\\x = \frac {5 \pm \sqrt {25-128}} {8}\\x = \frac {5 \pm \sqrt {-103}} {8}

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x = \frac {5 \pm \sqrt {103i ^ 2}} {8}\\x = \frac {5 \pm i \sqrt {103}} {8}

We have two imaginary roots:

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Answer:

x_ {1} = \frac {5 \ + i \sqrt {103}} {8}\\x_ {2} = \frac {5 \ -i \sqrt {103}} {8}

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