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2 years ago

3. Which of the following is an exponential function?

Mathematics

1 answer:

lukranit [14]2 years ago

4 0

Answer:

B is correct I hope this helps.

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What is the most precise name for quadrilateral ABCD with vertices A(-4, -4), B(-4, -2), C(-1, -2), and D(-1, -4)?

Darya [45]

Square Is the most precise name

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3 years ago

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Let V be the volume of the solid obtained by rotating about the y-axis the region bounded y = 4x and y = x2/4 . Find V by slicin

aliya0001 [1]

Answer:

Step-by-step explanation:

Consider the graphs of the $y = 4x$ and $y = \frac{x^{2} }{4}$ .

By equating the expressions, the intersection points of the graphs can be found and in this way delimit the area that will rotate around the Y axis.

$4x = \frac{x^{2} }{4} \\ x^{2} = 16x \\ x^{2} - 16x = 0 \\ x(x-16) = 0$ then $x=0$ o $x=16$ . Therefore the integration limits are:

$y = 4(0) = 0$ and $y = 4(16) = 64$

The inverse functions are given by:

$x = 2 \sqrt{y}$ and $x = \frac{y}{4}$ . Then

The volume of the solid of revolution is given by:

$\int\limits^{64}_ {0} \, [2\sqrt{y} - \frac{y}{4}]^{2} dy = \int\limits^{64}_ {0} \, [4y - y^{3/2} + \frac{y^{2}}{16} ]\ dy = [2y^{2} - \frac{2}{5}y^{5/2} + \frac{y^{3}}{48} ]\limits^{64}_ {0} = 546.133 u^{2}$

6 0

4 years ago

The graph below shows the solution to which set of inequalities?

fiasKO [112]

I think it is b but im not sure

7 0

3 years ago

Read 2 more answers

Two distinct number cubes, one red and one blue, are rolled together. Each number cube has sides numbered 1 through 6.

cestrela7 [59]

Answer:

$P(E_1 or E_2) = \frac{7}{12}$

Step-by-step explanation:

Given

Two cubes of side 1 - 6

Required

Probability that the outcome of the roll is an odd sum or a sum that is a multiple of 5

First, the sample space needs to be listed;

Let $C_r$ represent the Red cube

$C_b$ represent the Blue cube

S represent the sample space

$C_r = (1,2,3,4,5,6)\\C_b = (1,2,3,4,5,6)\\S = (2,3,4,5,6,7,3,4,5,6,7,8,4,5,6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12)$ S is gotten by getting the sum of $C_r$ and $C_b$

$n(S) = 36$

<em>Calculating the Probability</em>

Let $E_1$ represent the event that an outcome is an odd sum

$E_1 = (3,5,7,3,5,7,5,7,9,5,7,9,7,9,11,7,9,11)$

$n(E_1) = 18$

$P(E_1) = \frac{n(E_1)}{n(S)}$

$P(E_1) = \frac{18}{36}$

Let $E_2$ represent the event that an outcome is a multiple of 5

$E_2 = (5,5,5,5,10,10,10)$

$n(E_2) = 7$

$P(E_2) = \frac{n(E_2)}{n(S)}$

$P(E_2) = \frac{7}{36}$

Let $E_3$ represent the event that an outcome is an odd sum and a multiple of 5

$E_3 = E_1 and E_2$

$E_3 = (5,5,5,5)$

$n(E_3) = 4$

$P(E_3) = \frac{n(E_3)}{n(S)}$

$P(E_3) = \frac{4}{36}$

Calculating $P(E_1 or E_2)$

$P(E_1 or E_2) = P(E_1) + P(E_2) - P(E_1 and E_2)$

$P(E_1 or E_2) = P(E_1) + P(E_2) - P(E_3)$

$P(E_1 or E_2) = \frac{18}{36} + \frac{7}{36} - \frac{4}{36}$

$P(E_1 or E_2) = \frac{18 + 7 - 4}{36}$

$P(E_1 or E_2) = \frac{21}{36}$

$P(E_1 or E_2) = \frac{7}{12}$

Hence, the probability that the outcome of the roll is an odd sum or a sum that is a multiple of 5 is $\frac{7}{12}$

6 0

3 years ago

the air in a small room 12ft by 8ft by 8ft is 3% carbon monoxide. Starting at t=0, fresh air containing no carbon monoxide is bl

Irina-Kira [14]

Answer:

The air in the room at 0.01% carbon monoxide at 43.8 min

Step-by-step explanation:

Let be the volume of CO in the room at time t, be v(t) and the total volume of the room be V. The volume percent of CO in the room at a given time is then given by:

$p(t) = \frac{100\times v(t)}{V}$

Volume percent is the measure of concentration used in this problem. The "Amount" of CO in the room is then measured in terms of the volume of CO in the room.

Let the rate at which fresh air enters the room be f, which is the same as the rate at which air exits the room. We assume that the air in the room mixes instantaneously with the air entering the room, so that the concentration of CO is uniform throughout the room.

As you wrote, the rate at which the volume of CO in the room changes with time is given by

$\frac{dvt}{dt} = 0 \times f -\frac{f}{v} \times v(t) = -\frac{f}{v} \times v(t)$