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2 years ago

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A 2-column table with 8 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2, 3, 4. The

second column is labeled f of x with entries negative 2, 0, 2, 2, 0, negative 8, negative 10, negative 20. Which could be the entire interval over which the function, f(x), is positive? (–∞, 1) (–2, 1) (–∞, 0)

2 answers:

mrs_skeptik [129]2 years ago

5 0

Answer:

Option B.

Step-by-step explanation:

The given table of values is

x f(x)

-3 -2

-2 0

-1 2

0 2

1 0

2 -8

3 -10

4 -20

We need to find the interval for which the function f(x) is positive.

From the given table it is clear that the value of function f(x) is negative before -2 and after 1.

The function positive between x=-2 and x=1. So, we can conclude that the function f(x) is positive for the interval (-2,1).

Therefore, he correct option is B.

diamong [38]2 years ago

4 0

Answer:

it's b my friends ;)

Step-by-step explanation:

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you own a pet supply store, and the total sales in your trading area equal 27,000. Your market share is 45%. What were your sale

Mazyrski [523]

Answer:

The sales is Rs. 12150.

Step-by-step explanation:

Given : You own a pet supply store, and the total sales in your trading area equal 27,000. Your market share is 45%.

To find : What were your sales ?

Solution :

Total sales = 27,000

Market share = 45%

Sales is given by,

Sales is 45% of total sale.

So, $S=27000\times \frac{45}{100}$

$S=12150$

Therefore, the sales is Rs. 12150.

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2 years ago

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Let D be the smaller cap cut from a solid ball of radius 8 units by a plane 4 units from the center of the sphere. Express the v

natima [27]

Answer:

Step-by-step explanation:

The equation of the sphere, centered a the origin is given by $x^2+y^2+z^2 = 64$ . Then, when z=4, we get

$x^2+y^2= 64-16 = 48$ .

This equation corresponds to a circle of radius $4\sqrt[]{3}$ in the x-y plane

c) We will use the previous analysis to define the limits in cartesian and polar coordinates. At first, we now that x varies from $-4\sqrt[]{3}$ up to $4\sqrt[]{3}$ . This is by taking y =0 and seeing the furthest points of x that lay on the circle. Then, we know that y varies from $-\sqrt[]{48-x^2}$ and $\sqrt[]{48-x^2}$ , this is again because y must lie in the interior of the circle we found. Finally, we know that z goes from 4 up to the sphere, that is , z goes from 4 up to $\sqrt[]{64-x^2-y^2}$

Then, the triple integral that gives us the volume of D in cartesian coordinates is

$\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx$ .

b) Recall that the cylindrical coordinates are given by $x=r\cos \theta, y = r\sin \theta,z = z$ , where r corresponds to the distance of the projection onto the x-y plane to the origin. REcall that $x^2+y^2 = r^2$ . WE will find the new limits for each of the new coordinates. NOte that, we got a previous restriction of a circle, so, since $\theta[\tex] is the angle between the projection to the x-y plane and the x axis, in order for us to cover the whole circle, we need that [tex]\theta$ goes from 0 to $2\pi$ . Also, note that r goes from the origin up to the border of the circle, where r has a value of 4\sqrt[]{3}. Finally, note that Z goes from the plane z=4 up to the sphere itself, where the restriction is \sqrt[]{64-r^2}. So, the following is the integral that gives the wanted volume

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta$ . Recall that the r factor appears because it is the jacobian associated to the change of variable from cartesian coordinates to polar coordinates. This guarantees us that the integral has the same value. (The explanation on how to compute the jacobian is beyond the scope of this answer).

a) For the spherical coordinates, recall that $z = \rho \cos \phi, y = \rho \sin \phi \sin \theta, x = \rho \sin \phi \cos \theta$ . where $\phi$ is the angle of the vector with the z axis, which varies from 0 up to pi. Note that when z=4, that angle is constant over the boundary of the circle we found previously. On that circle. Let us calculate the angle by taking a point on the circle and using the formula of the angle between two vectors. If z=4 and x=0, then y=4 $\sqrt[]{3}$ if we take the positive square root of 48. So, let us calculate the angle between the vector $a=(0,4\sqrt[]{3},4)$ and the vector b =(0,0,1) which corresponds to the unit vector over the z axis. Let us use the following formula

$\cos \phi = \frac{a\cdot b}{||a||||b||} = \frac{(0,4\sqrt[]{3},4)\cdot (0,0,1)}{8}= \frac{1}{2}$

Therefore, over the circle, $\phi = \frac{\pi}{3}$ . Note that rho varies from the plane z=4, up to the sphere, where rho is 8. Since $z = \rho \cos \phi$ , then over the plane we have that $\rho = \frac{4}{\cos \phi}$ Then, the following is the desired integral

$\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{\frac{4}{\cos \phi}}^{8}\rho^2 \sin \phi d\rho d\phi d\theta$ where the new factor is the jacobian for the spherical coordinates.

d ) Let us use the integral in cylindrical coordinates

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta=\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} r (\sqrt[]{64-r^2}-4) dr d\theta=\int_{0}^{2\pi} d \theta \cdot \int_{0}^{4\sqrt[]{3}}r (\sqrt[]{64-r^2}-4)dr= 2\pi \cdot (-2\left.r^{2}\right|_0^{4\sqrt[]{3}})\int_{0}^{4\sqrt[]{3}}r \sqrt[]{64-r^2} dr$

Note that we can split the integral since the inner part does not depend on theta on any way. If we use the substitution $u = 64-r^2$ then $\frac{-du}{2} = r dr$ , then

$=-2\pi \cdot \left.(\frac{1}{3}(64-r^2)^{\frac{3}{2}}+2r^{2})\right|_0^{4\sqrt[]{3}}=\frac{320\pi}{3}$

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OleMash [197]

Step-by-step explanation:

given that 1.5 gram of fat =3% of 2000 calorie diet.

Thus, 1.5 gram of fat contributes to 3/100*2000= 60 calorie of diet.

Generally 6% of the diet must be fat.(recommended)

Since, 1.5 gram = 3%

thus, 3 gram =6%.

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riadik2000 [5.3K]

What is the perimeter?

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2 years ago

This is harder so i put more points

Inessa [10]

Answer:

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Step-by-step explanation:

7.6x - 2.3x = 106

We are finding the value of x.

5.3x = 106

Divide both sides by 5.3.

x = 106 ÷ 5.3

x = 20

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1 year ago

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