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Nady [450]
2 years ago
5

What is the value of b and the missing symbol in Jamilla’s inequality?

Mathematics
2 answers:
Vikentia [17]2 years ago
5 0

Answer:

The answer for my question was also C..

Step-by-step explanation:

Sindrei [870]2 years ago
4 0

Answer:

b=1, \geq

Step-by-step explanation:

we know that

The absolute value function has two solutions

Observing the graph

the solutions are

x\geq 1 and x\leq -3

First solution (case positive)

assume the symbol of the first solution and then compare the results

\left|x+b\right|\ge2

(x+b)\ge2

x\ge2-b

2-b=1\\b=2-1\\ b=1

Second solution (case negative)

-(x+b)\ge2

Multiply by -1 both sides

(x+b)\leq-2

substitute the value of b and compare the results

(x+1)\leq-2

x\leq-2-1

x\leq-3 -------> is correct

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S = d/t

st = d

t = d/s

The time going is t1.
The time returning is t2.
The total time is 4 hours, so we have t1 + t2 = 4

The speed of the current is c.
The speed going is 9 + c.
The speed returning is 9 - c.

t1 = 16/(9 + c)

t2 = 16/(9 - c)

t1 + t2 = 16/(9 + c) + 16/(9 - c)

4 = 16/(9 - c) + 16/(9 + c)

1 = 4/(9 - c) + 4/(9 + c)

(9 + c)(9 - c) = 4(9 - c) + 4(9 + c)

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81 - c^2 = 72

c^2 = 9

c^2 - 9 = 0

(c + 3)(c - 3) = 0

c + 3 = 0   or   c - 3 = 0

c = -3   or   c = 3

We discard the negative answer, and we get c = 3.

The speed of the current is 3 mph.
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A weight lifter can deadlift 275 pounds. She can increase the weight that she can lift according to the function W(x)=275(1.05)^
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350.98\ pounds

Step-by-step explanation:

we have

W(x)=275(1.05^{x})

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Consider a tank in the shape of an inverted right circular cone that is leaking water. The dimensions of the conical tank are a
Artyom0805 [142]

Answer:

\frac{dh}{dt} = \frac{216}{1875\pi}

Step-by-step explanation:

Given

Represent radius with r and height with h

h = 12ft

r = 10ft

Rate = \frac{8ft^3}{min} when h = 10ft

Express radius in terms of height

\frac{r}{h} = \frac{10}{12}

\frac{r}{h} = \frac{5}{6}

r = \frac{5}{6}h

First, we need to determine the volume of the cone in terms of height.

Volume = \frac{1}{3}\pi r^2h

Substitute r = \frac{5}{6}h

V = \frac{1}{3} * \pi * (\frac{5}{6}h)^2 * h

V = \frac{1}{3} * \pi * \frac{25}{36}h^2 * h

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\frac{dV}{dt} = 3 *  \frac{25}{108} \pi h^{3-1} * \frac{dh}{dt}

\frac{dV}{dt} = \frac{75}{108} \pi h^{2} \frac{dh}{dt}

Solve for \frac{dh}{dt}

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So, we have:

\frac{dh}{dt} = 8 * \frac{108}{75\pi 10^2}

\frac{dh}{dt} = \frac{8 * 108}{75\pi 10^2}

\frac{dh}{dt} = \frac{864}{7500\pi}

\frac{dh}{dt} = \frac{216}{1875\pi}

Hence:

The rate is \frac{216}{1875\pi} ft/min

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