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3 years ago

Examine the following system of inequalities.

Mathematics

1 answer:

lubasha [3.4K]3 years ago

7 0

Answer:

Dotted linear inequality shaded above passes through (0, 4) and (4, 0). Solid exponential inequality shaded below passes through (negative 2,2) & (0,5)

Step-by-step explanation:

we have

$y > -x+4$ ----> inequality A

The solution of the inequality A is the shaded area above the dotted line $y=-x+4$

The dotted line passes through the points (0,4) and (4,0) (y and x-intercepts)

and

$y \leq -(1/2)^{x} +6$ -----> inequality B

The solution of the inequality B is the shaded area above the solid line $y=-(1/2)^{x} +6$

The solid line passes through the points (0,5) and (-2,2)

therefore

The solution of the system of inequalities is the shaded area between the dotted line and the solid line

see the attached figure

Dotted linear inequality shaded above passes through (0, 4) and (4, 0). Solid exponential inequality shaded below passes through (negative 2,2) & (0,5)

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an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a

viktelen [127]

Answer:

the rate of change of the water depth when the water depth is 10 ft is; $\mathbf{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at a time t and r = radius of the cone shaped at a time t

Then the similar triangles ΔOCD and ΔOAB is as follows:

$\dfrac{h}{r}= \dfrac{20}{8}$ ( similar triangle property)

$\dfrac{h}{r}= \dfrac{5}{2}$

$\dfrac{h}{r}= 2.5$

h = 2.5r

$r = \dfrac{h}{2.5}$

The volume of the water in the tank is represented by the equation:

$V = \dfrac{1}{3} \pi r^2 h$

$V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h$

$V = \dfrac{1}{18.75} \pi \ h^3$

The rate of change of the water depth is :

$\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

Since the water is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

$\dfrac{dv}{dt}= - 4 \ ft^3/sec$

Therefore,

$-4 = \dfrac{\pi r^2}{6.25}\ \dfrac{dh}{dt}$

the rate of change of the water at depth h = 10 ft is:

$-4 = \dfrac{ 100 \ \pi }{6.25}\ \dfrac{dh}{dt}$

$100 \pi \dfrac{dh}{dt} = -4 \times 6.25$

$100 \pi \dfrac{dh}{dt} = -25$

$\dfrac{dh}{dt} = \dfrac{-25}{100 \pi}$

Thus, the rate of change of the water depth when the water depth is 10 ft is; $\mathtt{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

4 0

3 years ago

Tom bought a soft drink for $4.00 and 7 candy bars. He spent a total of $18.00. How

olga2289 [7]

Answer:

Step-by-step explanation:

$18-$4=$14

$14/7=$2

8 0

3 years ago

Solve for x: 4 over x plus 4 over quantity x squared minus 9 equals 3 over quantity x minus 3

dezoksy [38]

Answer:

Step-by-step explanation:

4/x + 4/(x²-9) = 3/(x - 3)

4 / x + 4 / [( x - 3) ( x + 3 )] = 3 / ( x - 3 ) / * x ( x - 3 ) ( x + 3 )

Restrictions : x ≠ 0, x ≠ - 3 , x ≠ 3;

4 ( x + 3 ) ( x - 3 ) + 4 x = 3 x ( x + 3 )

4 ( x² - 9 ) + 4 x = 3 x² + 9 x

4 x² - 36 + 4 x - 3 x² - 9 x = 0

x² - 5 x - 36 = 0

x² - 9 x + 4 x - 36 = 0

x ( x - 9 ) + 4 ( x - 9 ) = 0

( x - 9 ) ( x + 4 ) = 0

x - 9 = 0, or : x + 4 = 0

Answer:

x = 9, x = - 4

4 0

3 years ago

A school has 1800 students and 1800 light bulbs, each with a pull cord and all in a row. All the lights start out off. The first

Gnom [1K]

<u>Solution-</u>

A school has 1800 students and 1800 light bulbs, each with a pull cord and all in a row.

As all the lights start out off, in the first pass all bulbs will be turned on.

In the second pass all the multiples of 2 will be off and rest will be turned on.

In the third pass all the multiples of 3 will be off, but the common multiple of 2 and 3 will be on along with the rest. i.e all the multiples of 6 will be turned on along with the rest.

In the fourth pass 4th light bulb will be turned on and so does all the multiples of 4.

But, in the sixth pass the 6th light bulb will be turned off as it was on after the third pass.

This pattern can observed that when a number has odd number of factors then only it can stay on till the last pass.

1 = 1

2 = 1, 2

3 = 1, 3

<u>4 = 1, 2, 4</u>

5 = 1, 5

6 = 1, 2, 3, 6

7 = 1, 7

8 = 1, 2, 4, 8

9 = 1, 3, 9

10 = 1, 2, 5, 10

11 = 1, 11

12 = 1, 2, 3, 4, 6, 12

13 = 1, 13

14 = 1, 2, 7, 14

15 = 1, 3, 5, 15

16 = 1, 2, 4, 8, 16

so on.....

The numbers who have odd number of factors are the perfect squares.

So calculating the number of perfect squares upto 1800 will give the number of light bulbs that will stay on.

As, $\sqrt{1800} =42.42$ , so 42 perfect squared numbers are there which are less than 1800.

∴ 42 light bulbs will end up in the on position. And there position is given in the attached table.

7 0

3 years ago

The bottom of a ladder must be placed 3 ft from a wall. the ladder is 12 feet long. how far above the ground does the ladder tou

3241004551 [841]

9 feet above the ground, since it has to be 3 feet away from the wall you would have to subtract those 3 feet from the length of the ladder.

6 0

3 years ago