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3 years ago

State the domain and range of the following function.

Mathematics

1 answer:

insens350 [35]3 years ago

6 0

Hold on need you omgg disnwkxjdjcjdjxuj ex djdusijddj

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12/19*.............................= -12/19

melamori03 [73]

Answer:

$\frac{12}{19} \times \underline{\bold{-1}} = -\frac{12}{19}$

Step-by-step explanation:

$\frac{12}{19} \times ..... = -\frac{12}{19}$

= $\frac{\frac{12}{19}}{-\frac{12}{19}}$

= $\frac{12}{19} \times -\frac{19}{12}$

= $-\frac{228}{228}$

= $\bold{-1}$

So, the appropriate value to complete it is -1

#CMIIW ^^

3 0

3 years ago

A 5-meter ladder is leaning against the side of a house. The foot of the ladder is pulled away from the house at a rate of 0.4 m

Flura [38]

<h2>The top of the ladder is descending at 0.3 m/s.</h2>

Step-by-step explanation:

By Pythagoras theorem we know that

Hypotenuse² = Base² + Perpendicular²

h² = b² + p²

We have for ladder

h = 5 m

b = 3 m

5² = 3² + p²

p = 4 m

$\frac{db}{dt}=0.4m/s\\\\\frac{dh}{dt}=0$

Differentiating h² = b² + p² with respect to time

$2h\times \frac{dh}{dt}=2b\times \frac{db}{dt}+2p\times \frac{dp}{dt}\\\\5\times 0=3\times 0.4+4\times \frac{dp}{dt}\\\\\frac{dp}{dt}=-0.3m/s$

The top of the ladder is descending at 0.3 m/s.

3 0

3 years ago

What is the b in this math question -17+b/8=1

brilliants [131]

Are you supposed to find B?

5 0

3 years ago

The population of a town grows at a rate proportional to the population present at time t. The initial population of 500 increas

Mazyrski [523]

Answer:

The population in 40 years will be 1220.

Step-by-step explanation:

The population of a town grows at a rate proportional to the population present at time t.

This means that:

$P(t) = P(0)e^{rt}$

In which P(t) is the population after t years, P(0) is the initial population and r is the growth rate.

The initial population of 500 increases by 25% in 10 years.

This means that $P(0) = 500, P(10) = 1.25*500 = 625$

We apply this to the equation and find t.

$P(t) = P(0)e^{rt}$

$625 = 500e^{10r}$

$e^{10r} = \frac{625}{500}$

$e^{10r} = 1.25$

Applying ln to both sides

$\ln{e^{10r}} = \ln{1.25}$

$10r = \ln{1.25}$

$r = \frac{\ln{1.25}}{10}$

$r = 0.0223$

$P(t) = 500e^{0.0223t}$

What will be the population in 40 years

This is P(40).

$P(t) = 500e^{0.0223t}$

$P(40) = 500e^{0.0223*40} = 1220$

The population in 40 years will be 1220.

7 0

3 years ago

Your principal surprises you by buying your class a turtle. He brings the turtle to your class along with a sheet from the pet s

Nitella [24]

The height of the tank must be at least 1 foot, or 12 inches. We know the floor area (which is length x width) must be at least 400 inches. Therefore these minimum dimensions already tell us that the minimum volume is 400 x 12 = 4800 cubic inches. Since we have a maximum of 5000 cubic inches, the volume must be within the range of 4800 - 5000 cubic inches.

We can set the height at exactly 1 ft (or 12 inches). Then we can select length and width that multiply to 400 square inches, for example, L = 40 inches and W = 10 in. This gives us a tank of dimensions 40 x 10 x 12 = 4800 cubic inches, which fits all the criteria.

7 0

3 years ago