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4 years ago

3(3x+2)<8(x+2)<6(5x-12) =

2 answers:

8 0

$3(3x+2) < 8(x+2) < 6(5x-12)\\\\9x+6 < 8x+16 < 30x-72\\\\\#1\\9x+6 < 8x+16\ \ \ \ \ |subtract\ 6\ from\ both\ sides\\9x < 8x+10\ \ \ \ \ \ \ |subtract\ 8x\ from\ both\ sides\\x < 10\\\\\#2\\8x+16< 30x-72\ \ \ \ \ \ \ \ |subtract\ 16\ from\ both\ sides\\8x < 30x-88\ \ \ \ \ \ \ \ \ |subtract\ 30x\ from\ both\ sides\\-22x < -88\\22x > 88\ \ \ \ \ \ \ \ \ |divide\ both\ sides\ by\ 22\\x > 4\\\\From\ \#1\ \cap\ \#2\ we\ have\ 4 < x < 10\Rightarrow x\in(4;\ 10)$

kolezko [41]4 years ago

7 0

3(3x+2) < 8x + 16 => x < 10
8x + 16 < 30x - 72 => 22x > 88 => x > 4
solution is 4 < x < 10

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3 years ago

Comprás 4 libras de arroz con 10$.¿Cuántos necesitas para 14 libras?

Fofino [41]

Answer:

Alegría puede comprar 16.667 libras de arroz y 12.5 libras de azúcar con 20 pesos.

Step-by-step explanation:

Sabemos que una libra de arroz cuesta 0.60 pesos, mientras que una de azúcar cuesta 0.80 pesos. Un enfoque posible es determinar cuantas libras de arroz se obtiene por el mismo dinero para adquirir una libra de azúcar a través de una regla de tres simple:

Esto quiere decir que por cada tres libras de azúcar, se compra cuatro libras de arroz. Ahora, la cantidad de alimentos que puede comprar Alegría queda restringida al dinero disponible (20 pesos) y se describe por la siguiente ecuación:

(Ec. 1)

Donde:

- Cantidad de arroz, medida en libras.

- Cantidad de azúcar, medida en libras.

Además, la siguiente ecuación se deriva de la regla de tres del inicio del problema:

(Ec. 2)

Si aplicamos (Ec. 2) en (Ec. 1) y resolvemos la ecuación resultante, tenemos que:

De (Ec. 2) tenemos que la cantidad de arroz es:

En consecuencia, Alegría puede comprar 16.667 libras de arroz y 12.5 libras de azúcar con 20 pesos.

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3 years ago

Need help with math.

nikdorinn [45]

4 / (1/4 - 5/2)

4 / (1/4 -10/4)

4 / (-9/4)

4 x (-4/9) = -16/9

8 0

3 years ago

For each of the following expressions rewrite it in a way that would make it easier to simplify. State whether you used the asso

Elis [28]

1. 15(10 - 2)....distribute....150 - 30 = 120
2. (24 + 18) + (12 + 6).....commutative....(24 + 6) + (12 + 18) = 30 + 30 = 60
3. 5 * 13 * 2.....commutative.....5 * 2 * 13 = 10 * 13 = 130
4. 94 + 17 + 53 + 6....commutative...94 + 6 + 17 + 53 = 100 + 70 = 170
5. 25(14) ...distribute..25 * 10 + 25 * 4 = 250 + 100 = 350
6. 25 * (4 * 17)...associative....(25 * 4) * 17 = 100 * 17 = 1700

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4 years ago

Determine if diverges, converges, or converges conditionally.

TEA [102]

The given series is conditionally convergent. This can be obtained by using alternating series test first and then comparing the series to the harmonic series.

<h3>Determine if diverges, converges, or converges conditionally:</h3>

Initially we need to know what Absolute convergence and Conditional convergence,

If $\sum|a_{n} |$ → converges, and $\sum a_{n}$ → converges, then the series is Absolute convergence

If $\sum|a_{n} |$ → diverges, and $\sum a_{n}$ → converges, then the series is Conditional convergence

First use alternating series test,

$\lim_{k \to \infty} \frac{k^{5} +1}{k^{6}+11 }$ = $\lim_{n \to \infty} \frac{5}{6k}$ = 0,

The series is a positive, decreasing sequence that converges to 0.

Next by comparing the series to harmonic series,

$\sum^{\infty} _{k=2}|(-1)^{k+1} \frac{k^{5} +1}{k^{6}+11 }|=\sum^{\infty} _{k=2}\frac{k^{5} +1}{k^{6}+11 }$ ≈ $\sum^{\infty} _{k=2}\frac{1}{k}$ = 0

This implies that the series is divergent by comparison to the harmonic series.

First we got that the series is converging and then we got the series is divergent. Therefore the series is conditionally convergent.

$\sum|a_{n} |$ → diverges, and $\sum a_{n}$ → converges, then the series is Conditional convergence.

Hence the given series is conditionally convergent.

Learn more about conditionally convergent here:

brainly.com/question/1580821

#SPJ1

7 0

2 years ago

Read 2 more answers