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Murljashka [212]

2 years ago

13

Stacey owns a lot that has 180 feet of front footage and contains 36,000 square feet. She purchases two lots adjacent to each si

de of his lot. These side lots are each 200 feet deep and contain 19,000 square feet. What is the total front footage of all three lots?

2 answers:

vlabodo [156]2 years ago

7 0

<h2>The total front footage of all three lots is 370 ft</h2>

Step-by-step explanation:

Front footage of first lot = 180 ft

Area of each side lot = 19000 ft²

Depth of each side lot = 200 ft

We have

Area = Depth x Front footage

19000 = 200 x Front footage

Front footage = 95 ft

Total front footage = Front footage of first lot + 2 x Front footage of each side lot

Total front footage = 180 + 2 x 95

Total front footage = 180 + 190

Total front footage = 370 ft

The total front footage of all three lots is 370 ft

bija089 [108]2 years ago

3 0

The graph of the function f(x)=(x+2)(x+6) is shown

The statement the function is negative for all real values of x where x <-2

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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

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

3 0

2 years ago

Follow this link to view Jamie’s work. Critique Jamie’s work by explaining the reasonableness of the scenario and the accuracy o

Stella [2.4K]

Answer:

See below

Step-by-step explanation:

This graph could describe a relay race.

Jamie writes a scenario that can be modeled by the piecewise function on the graph.

The graph has time, in minutes on the x-axis and the distance, in miles on the y-axis.

The first part of the graph is the line starting from the origin through to the point (12, 1) then the second part starts from the point (12, 1) and ends on the point (20, 2).

1. A runner begins the race and runs at a steady pace for 1 minute.

This is incorrect as describes the wrong axis.

It should say: a runner begins a race and runs at a steady pace for 12 minutes, completing 1 mile distance.

<u>The equation for this part:</u>

y = mx + b
b = 0 as it starts at origin.
m = 1/12
y = 1/12x
x = [0, 12], y = [0, 1]

2. His partner then takes over and runs at a steady, but faster, pace for another 1 minute.

Again, as above, wrong axis is described.

It should say: his partner then takes over and runs at a steady, but faster, pace for another 8 minutes, completing the last 1 mile of the race.

<u>The equation for this part:</u>

b =1, as a starting point
m = (2-1)/(20-12) = 1/8
y = 1/8x + 1
x = (12, 20], y = (1, 2]

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2 years ago

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myrzilka [38]

Answer:

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Step-by-step explanation:

Given that a person has 12000 rupees for his daily expenses.

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Given that number of day is increased by 2 more days, that is number of days is x+2.

New daily expense per day $\frac{12000}{x+2}$

Given that this new daily expenses are 300 less than original.

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$\frac{12000(x+2)-12000x}{x(x+2)}=300$

$\frac{24000}{x(x+2)}=300$

$x(x+2)=\frac{24000}{300}$

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x=8 or -10.

Since number of days cannot be negative, duration of the tour planned=8.

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