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Svet_ta [14]
1 year ago
9

Which graph represents the solution set of the compound inequality? −5 < a − 4 < 2

Mathematics
2 answers:
never [62]1 year ago
8 0

-5 < a - 4 < 2\\\\ -1

the 3rd one

jeyben [28]1 year ago
3 0

Answer:

Correct graph is:

Third

Step-by-step explanation:

we have to determine the graph of the inequality:

−5 < a − 4 < 2

Adding 4 on each side of the above inequality:

-5+4 < a-4+4 < 2+4

i.e. -1 < a < 6

Hence, graph of the inequality −5 < a − 4 < 2 is the set of all values between -1 and 6 and excluding -1 and 6

Correct graph is:

Third

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How many 2/5 cup servings are in 9/5 cups of ice cream
Aneli [31]

Answer:

5/2

Step-by-step explanation:

5/2 because 2/5 times 4 is 8/2 plus 1/2 is 9/2

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2 years ago
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Data berikut ialah suhu yang dicatat untuk 5 hari.
murzikaleks [220]

\frac{25 + 2 7 + 34 + 28 + 29}{5}

119.8

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1 year ago
La fuerza necesaria para evitar que un auto derrape en una curva varía inversamente al radio de la curva y conjuntamente con el
Vika [28.1K]

Answer:

768 libras de fuerza

Step-by-step explanation:

Tenemos que encontrar la ecuación que los relacione.

F = Fuerza necesaria para evitar que el automóvil patine

r = radio de la curva

w = peso del coche

s = velocidad de los coches

En la pregunta se nos dice:

La fuerza requerida para evitar que un automóvil patine alrededor de una curva varía inversamente con el radio de la curva.

F ∝ 1 / r

Y luego con el peso del auto

F ∝ w

Y el cuadrado de la velocidad del coche

F ∝ s²

Combinando las tres variaciones juntas,

F ∝ 1 / r ∝ w ∝ s²

k = constante de proporcionalidad, por tanto:

F = k × w × s² / r

F = kws² / r

Paso 1

Encuentra k

En la pregunta, se nos dice:

Suponga que 400 libras de fuerza evitan que un automóvil de 1600 libras patine alrededor de una curva con un radio de 800 si viaja a 50 mph.

F = 400 libras

w = 1600 libras

r = 800

s = 50 mph

Tenga en cuenta que desde el

F = kws² / r

400 = k × 1600 × 50² / 800

400 = k × 5000

k = 400/5000

k = 2/25

Paso 2

¿Cuánta fuerza evitaría que el mismo automóvil patinara en una curva con un radio de 600 si viaja a 60 mph?

F = ?? libras

w = ya que es el mismo carro = 1600 libras

r = 600

s = 60 mph

F = kws² / r

k = 2/25

F = 2/25 × 1600 × 60² / 600

F = 768 libras

Por lo tanto, la cantidad de fuerza que evitaría que el mismo automóvil patine en una curva con un radio de 600 si viaja a 60 mph es de 768 libras.

7 0
2 years ago
Sweettooth frozen yogurt has 12 different toppings to choose from. if they add two new topping, how many ways more can ian choos
tester [92]
Assuming that the topping order is not important, you need to use the combination to solve this question. The number of toppings is 12 and then added 2, so the number will become: 12+2= 14 toppings

From 14 toppings, ian need to choose 3. The possible ways would be:
14C3= 14!/(14-3)!3!= 14*13*12/ 3*2= 364 possible ways
7 0
1 year ago
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If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe
Natalka [10]

Answer:

They encircle the planet 3.76\times 10^{11} times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is 6.022\times 10^{23}

Thus, the number of dust particles is:

2.5\times 6.022\times 10^{23}=15.055\times 10^{23}

Diameter of a dust particles is 10μm and the circumference of earth is 40,076 km.

Convert the measurement in meters.

Diameter: 10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m

If we line up the particles the distance they could cover is:

15.055\times 10^{23}\times 10^{-5}=15.055\times 10^{18}=1.5055\times 10^{19}

Circumference in meters:

40,076km\times \frac{1000m}{1km}=40,076,000 m

Therefore,

\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}

Hence, they encircle the planet 3.76\times 10^{11} times.

8 0
2 years ago
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