Solving Two Step Equations

Of the students in the band, 1/6 play the flute and another 1/6 play the clarinet. What fraction of the students in the band pla

yaroslaw [1]

Answer:

2/6 because the 6 is the total and there is only one playing in each

Step-by-step explanation:

brainly please

7 0

3 years ago

Expand 5(8x - 3y) to its simplest form

Olenka [21]

Answer:

40x - 15y

Step-by-step explanation:

Multiply the brackets together

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Have a good day :)

6 0

3 years ago

Read 2 more answers

Without working out the multiplication, which product is larger, why? 3 × 1/5 or 3 × 1/10 A) 3 × 1/5 because 1/5 is less than 1/

sp2606 [1]

The answer to that question is b

6 0

3 years ago

The Sun is roughly 10^2 times as wide as the Earth. The Star KW Sagittarii is roughly 10^5 times as wide as the Earth. About how

Studentka2010 [4]

Answer:

1000 times

Step-by-step explanation:

Given:

The Sun is roughly 10^2 times as wide as the Earth.

The Star KW Sagittarii is roughly 10^5 times as wide as the Earth.

Question asked:

About how many times as wide as the Sun is KW Sagittarii?

Solution:

Let the width of the earth = $x$

As the Sun is roughly 10^2 times as wide as the Earth, hence the width of the sun = $x\times10^{2}$

And as the Star KW Sagittarii is roughly 10^5 times as wide as the Earth, hence the width of the Star = $x\times10^{5}$

Now, to find that how many times width of the Star KW Sagittarii is as respect to the width of the Sun, we will simply divide:

Width of the Star KW Sagittarii = $x\times10^{5}$

Width of the Sun = $x\times10^{2}$

$\frac{x\times10^{5} }{x\times10^{2} }$

x canceled by x

$\frac{10^{5} }{10^{2} } =10^{5-2} =10^{3} =10\times\ 10\times10=1000$

Therefore, Star KW Sagittarii is 1000 times wider than Sun.

<em>First of all we calculated width of Sun in terms of width of earth and then calculated the width of the Star in terms of earth and for comparison we did simple division that showed that the Star KW Sagittarii is 1000 times wider than the Sun.</em>

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4 0

3 years ago

De acuerdo con la tercera ley de movimiento planetario de Kepler, la masa de un planeta es directamente proporcional al cubo de

Sunny_sXe [5.5K]

Answer:

La masa del Sol es $2.509\times 10^{31}$ kilogramos.

Step-by-step explanation:

Tras una lectura cuidadosa al enunciado, tenemos que la Tercera Ley de Kepler queda descrita por la siguiente relación:

$M \propto \frac{r^{3}}{T^{2}}$

$M = k\cdot \frac{r^{3}}{T^{2}}$ (Eq. 1)

Donde:

$r$ - Distancia entre los centros del planeta y el satélite, medido en kilómetros.

$T$ - Período oribital del satélite, medido en días.

$k$ - Constante de proporcionalidad, medida en kilogramo-días cuadrados por kilómetro cúbico.

$M$ - Masa del planeta, medida en kilogramos.

Podemos obtener la masa del Sol mediante la siguiente relación:

$\frac{M_{S}}{M_{E}} = \frac{\frac{r_{E}^{3}}{T_{E}^{2}} }{\frac{r_{M}^{3}}{T_{M}^{2}} }$

$\frac{M_{S}}{M_{E}} = \left(\frac{T_{M}}{T_{E}} \right)^{2}\cdot \left(\frac{r_{E}}{r_{M}} \right)^{3}$ (Eq. 2)

Donde:

$T_{M}$ , $T_{E}$ - Períodos orbitales de la Luna y la Tierra, medidos en días.

$r_{E}$ , $r_{M}$ - Distancias entre la Tierra y el Sol, así como entre la Luna y la Tierra, medidas en kilómetros.

$M_{S}$ , $M_{E}$ - Masas del Sol y la Tierra, medidos en kilogramos.

Si $M_{E} = 75.97\times 10^{24}\,kg$ , $T_{E} = 365.3\,d$ , $T_{M} = 27.3\,d$ , $r_{M} = 3.84\times 10^{5}\,km$ y $r_{E} = 1.496\times 10^{8}\,km$ , entonces tenemos que la masa del Sol es:

$M_{S} = \left(\frac{T_{M}}{T_{E}} \right)^{2}\cdot \left(\frac{r_{E}}{r_{M}} \right)^{3}\cdot M_{E}$

$M_{S} = \left(\frac{27.3\,d}{365.3\,d} \right)^{2}\cdot \left(\frac{1.496\times 10^{8}\,km}{3.84\times 10^{5}\,km} \right)^{3}\cdot (75.97\times 10^{24}\,kg)$

$M_{S} = 2.509\times 10^{31}\,kg$

La masa del Sol es $2.509\times 10^{31}$ kilogramos.

7 0

3 years ago