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Oxana [17]
2 years ago
10

(04.06 MC) In the graph, the area below f(x) is shaded and labeled A, the area below g(x) is shaded and labeled B, and the area

where f(x) and g(x) have shading in common is labeled AB. Graph of two intersecting lines. One line f of x is solid and goes through the points 0, 4, and 4, 0 and is shaded above the line. The other line g of x is solid, and goes through the points 0, negative 1 and 2, 5 and is shaded in below the line. The graph represents which system of inequalities? y ≤ −3x − 1 y ≤ −x − 4 y ≥ −3x + 1 y ≤ −x − 4 y ≤ 3x − 1 y ≤ −x + 4 y ≤ 3x − 1 y ≥ −x + 4

Mathematics
1 answer:
KonstantinChe [14]2 years ago
6 0

Answer:

(D)y ≤ 3x − 1 y ≥ −x + 4

Step-by-step explanation:

The line f(x) is solid and goes through the points (0, 4) and (4, 0) and is shaded above the line.

The line that satisfies the point (0,4) and (4,0) is y=-x+4

Since it is shaded above the line, we have the inequality sign  ≥.

Therefore, one of the lines is y≥-x+4

In the equation of the line y=3x-1

When x=0, y=3(0)-1=-1

When x=2, y=3(2)-1=5

Since it is shaded below the line, we have the inequality sign ≤ .

Therefore, the other line is y≤3x-1.

The correct option is D.

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The distance to the surface of the water in a well can sometimes be found by dropping an object into the well and measuring the
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THE QUESTION IS NOT PROPERLY WRITTEN

The distance to the surface of the water in a well can sometimes be found by dropping an object into the well and measuring the time elapsed until a sound is heard. If t1 is the time (in seconds) it takes for the object to strike the water, then t1 obeys the equation s = 16t1², where s is the distance (in feet) Solving for t1, we have t1 = √s/4. Let t2 be the time it takes for the sound of the impact to reach your ears. Since sound waves travel at a speed of approximately 1100 feet per second, the time to travel the distance s is t2 = s/1100. Now t1 +t2 is the total time that elapses from the moment that the object is dropped to the moment that a sound is heard. We have the equation; Total elapsed time √s/4 + s/1100.

Find the distance to the water’s surface if the total time elapsed from dropping a rock to hearing it hit water is 4 seconds.

Answer:

229.94 feet

Step-by-step explanation:

Given

t1 = √s/4

t2 = s/1100

From the question, we understand that

t1 + t2 = 4 seconds

Let y = √s

So,

t1 + t2 = 4 becomes

√s/4 + s/1100 = 4

y/4 + y²/1100 = 4 ------ Solve the fractions

(275y + y²)/1100 = 4

275y + y² = 4 * 1100

275y + y² = 4400 ------- Rearrange

y² + 275y - 4400 = 0 (Quadratic Equation)

The standard form of a quadratic equation is ax² + bx + c = 0

Where x =

frac{-b+-\sqrt{bx^{2} - 4ac } }{2a}

Here a = 1, b = 275 and c = -4400

So,

y = (-275+- √(275² - 4*1*-4400))/2

y = (-275+- √75625 + 17600))/2

y = (-275+- √(93225))/2

y = (-275 +- 305.328)/2

y = (-275 + 305.328)/2 or (-275-305.328)/2

y = 30.328/2 or -580.328/2

y = 15.164 or -290.164 -------Negative is not applicable

So, y = 15.164

But y = √s

So, s = y²

S = 15.164²

s = 229.94 ft

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La fuerza necesaria para evitar que un auto derrape en una curva varía inversamente al radio de la curva y conjuntamente con el
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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.

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Step-by-step explanation:

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