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Damm [24]
3 years ago
10

Line g is dilated by a scale factor of one half from the origin to create line g'. Where are points E' and F' located after dila

tion, and how are lines g and g' related?
A. The locations of E' and F' are E' (−8, 0) and F' (0, 4), and lines g and g' intersect at point F.
B. The locations of E' and F' are E' (−4, 0) and F' (0, 2), and lines g and g' are the same line.
C. The locations of E' and F' are E' (−2, 0) and F' (0, 1), and lines g and g' are parallel.
D. The locations of E' and F' are E' (−1, 0) and F' (0, 0), and lines g and g' are not related.

Mathematics
2 answers:
Bingel [31]3 years ago
7 0

Answer: The locations of E' and F' are E' (−2, 0) and F' (0, 1), and lines g and g' are parallel.

Step-by-step explanation:

If we have the graph of the equation y = f(x)

A dilation from the origin by a given factor, means that we multilpicate the function by that factor, so our new graph is:

y' = a*f(x)

where a is the scale factor, here a = 1/2.

so g' = (1/2)*g

then if E = (-4, 0)

          E' = (1/2)*E = (-2, 0)

          F = (0, 2)

          F' = (1/2)*F = (1, 0)

Then the correct option is C The locations of E' and F' are E' (−2, 0) and F' (0, 1), and lines g and g' are parallel.

MatroZZZ [7]3 years ago
5 0

Answer:

C. The locations of E' and F' are E' (−2, 0) and F' (0, 1), and lines g and g' are parallel.

Step-by-step explanation:

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slavikrds [6]
Vamos lá. 

<span>Pede-se para determinar o parâmetro "m" da equação abaixo, sabendo-se que uma raiz é nula e a outra é positiva: </span>

<span>x² + mx + m² - m - 12 = 0 </span>

<span>Veja que se uma raiz é nula (é igual a zero), então vamos substituir o "x" por "0", na equação acima: </span>

<span>0² + m*0 + m² - m - 12 = 0 </span>
<span>0 + 0 + m² - m - 12 = 0 </span>
<span>m² - m - 12 = 0 ------resolvendo essa equação do 2º grau você encontrará as seguintes raízes: </span>

<span>m' = -3 </span>
<span>m'' = 4 </span>

<span>Dessa forma, vamos substituir "m" por (-3) e por 4 e ver se a equação terá uma raiz nula e outra positiva. Vamos ver? </span>

<span>Substituindo "m" por "-3", ficamos com: </span>

<span>x² - 3x + (-3)² - (-3) - 12 = 0 </span>
<span>x² - 3x + 9 + 3 - 12 = 0 </span>
<span>x² - 3x +12 - 12 = 0 </span>
<span>x² - 3x = 0 <------Veja que as raízes dessa equação são: x' = 0 e x'' = 3 </span>
<span>Veja que para m = -3, a equação se verifica, pois temos uma raiz igual a "0" e a outra positiva (igual a 3). </span>

<span>Agora vamos substituir "m" por 4 na equação original: </span>

<span>x² + 4x + 4² - 4 - 12 = 0 </span>
<span>x² + 4x + 16 - 16 = 0 </span>
<span>x² + 4x = 0 <----- Veja que as raízes dessa equação são: x' = 0 e x'' = -4. </span>
<span>Observe que, para m = 4, a equação NÃO se verifica, pois temos uma raiz igual a "0" e a outra negativa (igual a -4). E no enunciado é informado que uma raiz deverá ser nula e a outra positiva. Como deu uma nula e a outra negativa, então m = 4 não convém. </span>

<span>Logo, o valor de "m" deverá ser: </span>

<span>m = -3 <----Pronto. Essa é a resposta. </span>
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