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Naya [18.7K]

Answer:

1. The value of the variable, y is 11

2. (B) QRS is congruent to segment (E) ΔWXY by the hypotenuse leg congruency criteria

Step-by-step explanation:

1. The lengths of the sides of the given triangles ABC are AB = 14, BC = 27, AC = 19, and ∡A = 32°

The lengths of the sides of the given triangle FGH, FG = 14, GH = 19, FH = 2y + 5, ∡G = 32°

From the given parameters, we have;

Segment AB (AB = 14) is congruent to segment FG (FG = 14)

Segment AC (AC = 19) is congruent to segment GH (GH = 19)

Angle ∡A (∡A = 32°) is congruent to angle ∡G (∡G = 32°)

∴ ΔBAC is congruent to ΔFGH by the Side-Angle-Side rule of congruency

Therefore, segment BC is congruent to segment FH by Congruent Parts of Congruent Triangle are Congruent, CPCTC

Segment BC = Segment FH by definition of congruency

∴ 27 = 2·y + 5

2·y + 5 = 27

2·y = 27 - 5 = 22

y = 22/2 = 11

y = 11

The value of the variable, y = 11

2. For option A. the vertices of triangle ABC are A(-7, 4), B(-4, 1), C(-2, 5)

The length of the sides are;

The length of side AB = √((-4 - (-7))² + (1 - 4)²) = 3·√2

The length of side BC = √((-4 - (-2))² + (1 - 5)²) = √20

The length of side AC = √((-2 - (-7))² + (5 - 4)²) = √26

For option B. the vertices of triangle QRS are Q(3, -4), R(3, -1), S(7, -1)

The length of the sides are;

The length of side QR = √((3 - 3)² + ((-4) - (-1))²) = 3

The length of side RS = √((7 - 3)² + (-1 - (-1))²) = 4

The length of side QS = √((3 - 7)² + ((-4) - (-1))²) = 5

For option C. the vertices of triangle DEF are D(-2, 6), E(1, 3), F(3, 7)

The length of the sides are;

The length of side DE = √(((-2) - 1)² + (6 - 3)²) = 3·√2

The length of side EF = √((3 - 1)² + (7 - 3)²) = √20

The length of side DF = √((3 - (-2))² + (7 - 6)²) = √26

For option D. the vertices of triangle TUV are T(-6, -5), U(-6, 1), V(4, 1)

The length of the sides are;

The length of side TU = √(((-6) - (-6))² + ((-5) - 1)²) = 6

The length of side UV = √(((-6) - 4)² + (1 - 1)²) = 10

The length of side TV = √(((-6) - 4)² + ((-5) - 1)²) = 2·√34

For option E. the vertices of triangle WXY are W(-6, 4), X(-6, 1), Y(-2, 1)

The length of the sides are;

The length of side WX = √(((-6) - (-6))² + (4 - 1)²) = 3

The length of side XY = √(((-6) - (-2))² + (1 - 1)²) = 4

The length of side WY = √(((-6) - (-2))² + (4 - 1)²) = 5

Therefore;

Segment QR of ΔQRS is congruent to segment WX of ΔWXY

Segment RS of ΔQRS is congruent to segment XY of ΔWXY

Segment QS of ΔQRS is congruent to segment WY of ΔWXY

Whereby QS and WY are the hypotenuse side of ΔQRS and ΔWXY respectively, because QS = WY = 5 = √( $\overline {QR} ^2$ + $\overline {RS} ^2$ ) = (√(3² + 4²)

and also RS = XY, by the definition of congruency, we have;

QRS is congruent to segment ΔWXY by the hypotenuse leg congruency criteria

6 0

3 years ago

How do you write 5h+3k in a word form

RideAnS [48]

5 of a number plus 3 of a number

6 0

3 years ago

John and 3 friends are going out for pizza for lunch. They split one pizza and 4 large drinks. The pizza cost $13.50. The

Nuetrik [128]

Answer:

7 dollars

Step-by-step explanation:

do 20.50 minus 13.50 but you can just take out the .50 because both sides have them and they will cancel out anyways then its just 20 minus 13 and that is 7.

6 0

4 years ago

8. Let R be the relation on the set of all sets of real numbers such that SRT if and only if S and T have the same cardinality.

sergejj [24]

Answer:

We must prove that the relation is reflexive, symmetric and transitive. Recall that to sets have the same cardinality if there exist a bijective mapping between them.

Reflexive: Take the identity map $I:S\rightarrow S$ , which is bijective. Then SRS.

Symmetric: If SRT then, there exist a bijective map $f:S\rightarrow R$ . In order to prove that TRS just take the inverse map of $f$ : $f^{-1}$ which is also bijective. Therefore, TRS.

Transitivity: Suppose that SRT and TRU. Also, assume that $f$ is the bijective map between S and T, and $g$ the bijective map between T and U. It is not difficult to check that the map $h=g(f)$ is bijective and $h:S\rightarrow U$ . Therefore, SRU.

Hence, the relation R is an equivalence relation.

The equivalence class of the set {0,1,2} is the class of all the sets with three elements, and we can associate it with the number 3. There is a construction of natural numbers based on this idea.

The equivalence class of Z is the same equivalence class of N. Therefore, is the class of all denumerable or countable sets.

Step-by-step explanation:

When we want to prove that a given relation R is equivalence, we need to check that R satisfies all the three conditions: reflexive, symmetric and transitivity. Usually the first two are very simple to prove and comes directly from the definition. The transitivity is more tricky. In this case we need to recall the definition of cardinality.

7 0

3 years ago

You have $120 to buy food trays for a super bowl party. The vegetable trays cost $20 each and the sandwich trays cost $30 each.

monitta

Answer:

3 vegetable trays, 2 sandwich trays

Step-by-step explanation:

I'm not quite sure what we're supposed to find, but if we were to find how many of each we can buy with $120 it would be 3 vegetable trays and 2 sandwich trays.

20 + 20 + 20 = $60

30 + 30 = $60

60 + 60 = $120

6 0

3 years ago

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