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lara31 [8.8K]
3 years ago
8

a red string of holiday lights blinks once every 3 seconds while a string of blue lgihts blink once every 4 seconds. how many ti

mes with both sets of lights blink at the same time in 1 minute
Mathematics
1 answer:
gavmur [86]3 years ago
6 0
The highest common factor of 4 and 3 is 12
one minute has 60 secs
thus it will blink 60/12=5 times
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liraira [26]

Answer:

x>\frac{1}{2} \\x

Step-by-step explanation:

If we solve for 'x' variable, its recommended to multiply 'x' variable in both sides of inequality:

we have \frac{1}{4}< x^{2} \\

This is a quadratic equation, two answer are going to be obtained from here.

since \sqrt{x^{2} } = |x|

Applying square roots to both sides of inequality sign we wil have the following

\sqrt{\frac{1}{4} } < \sqrt{x^{2} }

This leaves to the following

\frac{1}{4}

remember that |x|= +/ - x

so

x>\frac{1}{2} \\x

6 0
3 years ago
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RSB [31]
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3 0
3 years ago
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Does anyone know who to calculate the area of the arrow? Need help, please show.
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3 years ago
(CESGRANRIO) Determine o parâmetro m na equação x²+mx+m²-m-12=0, de modo que ela tenha uma raíz nula e outra positiva.
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>
4 0
3 years ago
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