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2 years ago

What transformation is shown below? (Look carefully at the vertices.)

Mathematics

2 answers:

Orlov [11]2 years ago

7 0

Answer:

translation

Step-by-step explanation:

unless theres something else to it pretty sure its just Translation...

olasank [31]2 years ago

7 0

This is translation.

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Help please it is math

Alex

7x+100+68.25 Let me know if you want me to explain it to you

8 0

3 years ago

There are five identical blue books, two identical green books, and three identical black books. How many different patterns can

dsp73

Answer:

2520 patterns

Step-by-step explanation:

In 'n' 10! ways, books can be arranged. But, there are also 5! permutation of blue books 'n1', 2! permutation of identical green books 'n2', and 3! permutation identical black books 'n3'.

Therefore, for non identical arrangements:

$\frac{n!}{n1!n2!n3!}$

$\frac{10!}{5!2!3!}$ = 2520

Therefore, the books can be arranged on a shelf in 2520 patterns

8 0

3 years ago

How can the zero product be applied to real life

ddd [48]

Answer:

The zero product property states that if a⋅b=0 then either a or b equal zero. This basic property helps us solve equations like (x+2)(x-5)=0.

The Zero Product Property simply states that if ab=0 , then either a=0 or b=0 (or both). A product of factors is zero if and only if one or more of the factors is zero.

Step-by-step explanation:

hope it helps

7 0

2 years ago

For the middle school dance, the student council ordered 22 balloons printed with the school logo. On the night of the party, 6

77julia77 [94]

Answer:

ANSWER FOR ONE IS 6

AND THE TWO IS:7

4 0

2 years ago

Para reunir dinero para su gira de estudios , los alumnos de un curso deciden vender números de una rifa que se encuentran numer

QveST [7]

Respuesta:

0.53

Explicación:

Para calcular la posibilidad del evento A: "ganar la rifa comprando todos los números múltiplos de 3 o 5", debemos usar la siguiente fórmula.

P(A) = casos favorables / casos posibles

Evaluemos primero todos los casos que son múltiplos de 3, entre 1 y 100: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99. En total son 33.

Ahora, evaluemos todos los casos que son múltiplos de 3, entre 1 y 100: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100. En total son 20.

El número total de casos favorables es 33 + 20 = 53.

El número de casos posibles es el total de números de 1 a 100, es decir 100.

Luego P(A) = 53/100 = 0.53.

7 0

2 years ago