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3 years ago

5

el precio de xbox es de $4,300 pesos .a esta cantidada se debe agregar 16% del iva ¿cual es el precio del x-box con el iva inclu

ido

1 answer:

777dan777 [17]3 years ago

4 0

Answer: 4,988 porque 16% de 4,300 es 688 asi que nomas lo aggregas a lo que costaba y agaras el resultado de 4,988

Step-by-step explanation:

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PLEASE HELP!!!!!

leva [86]

Answer:

11.7 units

Step-by-step explanation:

sq root(2--4)^2+(-7-3)^2

sq root(6)^2 + (-10)^2

sq root (36)+(100)

sq root 136

equals 11.66 which rounds up to 11.7

6 0

2 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.

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

$xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}$

$xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}$

$xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)$

$\implies \boxed{xyz = x + y + z + 2}$

(C)

6 0

2 years ago

Sue's average score for three bowling games was 162. In the second game, Sue scored 10 less than in the first game. In the third

natali 33 [55]

Answer:

Her score in the first game was 173.

Step-by-step explanation:

Sue's total score for the 3 games = 3*162 = 486 points.

Let her score in the first game be x points. Then in the second game she scored (x - 10) and in the third, ( x - 10 - 13) points:

x + (x - 10) + (x - 23) = 486

3x - 33 = 486

3x = 519

x = 173 (answer).

7 0

2 years ago

Read 2 more answers

Even if I have use bidmas I'm still confuse with this equation <br> (-7i)(2+i)+(2i)(-7i)

zheka24 [161]

Assuming "i" is an imaginary unit:
$(-7i)(2+i)+(2i)(-7i)=\\
-14i+7+14=\\
21-14i$

5 0

3 years ago

What is the expanded form for 2.059

Artyom0805 [142]

2+ 0.0+ 0.05+ 0.009okay so you just have to add the numbers. so you add 2+0.0+0.5+0.009. then you will get you answer. i hope this helps!!!!

3 0

2 years ago