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3 years ago

Metodo igualacion A) x+ y = 12 B) x - y = 4 ?

1 answer:

6 0

<span>No sé la forma correcta de hacerlo, pero encontrar conjuntos de dos números que pueden igualar 12, luego encontrar cuál de los encaja en la segunda ecuación correctamente.</span>

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Hi I really need help with this question please,

Mrrafil [7]

Answer:

${ ( \sqrt{30 - x} )}^{2} = {x}^{2} \\ 30 - x = {x}^{2} \\ \\ {x}^{2} + x - 30 = 0 \\ delta = 1 - 4 \times ( - 30) \times 1 = 121 \\ x1 = \frac{ - 1 + \sqrt{121} }{2 \times 1} = \frac{ - 1 + 11}{2} = 5 \\ x2 = \frac{ - 1 - \sqrt{121} }{2 \times 1} = \frac{ - 1 - 11}{2} = - 6 \\$

6 0

2 years ago

mr yeo had 32litre of water.he used 3/8 of it to fill up a fish tank.he used 1/4 of the remaining water to water the plants. how

aleksandr82 [10.1K]

So, Mr Yeo has 32 litres of water to begin, by taking away the 3/8 (12 litres) you get 20 litres, dividing that by 4 gives you 5litres.

Your answer is then 5 Litres

5 0

3 years ago

Read 2 more answers

Wich expression is equivelint-3/4 divided by -8/12

Lisa [10]

Answer:

-1 1/8

Step-by-step explanation:

-3/4 / -8/12

= -3/4 * -12/8

= -3*12 / -4*8

= -36/32

= -1 4/32

= -1 1/8

7 0

3 years ago

If <img src="https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20x%20%3D%20log_%7Ba%7D%28bc%29" id="TexFormula1" title="\rm \: x = log_{a}(

timama [110]

Use the change-of-basis identity,

$\log_x(y) = \dfrac{\ln(y)}{\ln(x)}$

to write

$xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}$

Use the product-to-sum identity,

$\log_x(yz) = \log_x(y) + \log_x(z)$

to write

$xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}$

Redistribute the factors on the left side as

$xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}$

and simplify to

$xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)$

Now expand the right side:

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}$

Simplify and rewrite using the logarithm properties mentioned earlier.

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1$