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2 years ago

5

The parking spaces in a city lot were created by painting rows of parallel line segments, then painting a line that cuts through

each of the parallel segments. A sketch of two parking spaces is shown below. If m∠2 = 115°, what is m∠1?

2 answers:

Alona [7]2 years ago

5 0

Answer:

Step-by-step explanation: The answer is 65 degrees because the line in the middle is a 180 degree angle and 180-115=65. hope this helps :)

damaskus [11]2 years ago

4 0

the answer is 65 degrees because the line in the middle is a 180 degree angle and 180-115=65. hope this helps :)

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Sue’s Corner Market has a markup of 60% on bottled water. If the market sells a bottle of water for $2, find the original amount

lakkis [162]

Markup is the amount added to the cost price of goods to cover overhead and profit.

Sue’s Corner Market has a markup of 60% on bottled water.

Let us say original price was $x.

Now price after markup is $2.

So we can make an equation like:

original price + markup price = price after markup

x + 60% of x =2

$x + 0.6x =2$

$1.6x =2$

dividing both sides by 1.6

x= 1.25

So original price was 1.25 dollars.

5 0

2 years ago

Read 2 more answers

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

Jillian is selling boxes of cookies to raise money for her basketball team. The 10 oz. box costs $3.50, while the 16 oz. box cos

cluponka [151]

<span>we have to solve simultaneous equations.

x = 10 oz. boxes

y = 16 oz. boxes

simultaneous equations

x + y = 24 ---1)

3.50x + 5.00y = 97.50 ----2)

to solve for x we need to eliminate y, therefore we have to multiply 1st equation by 5

5x + 5y = 120 --3)

subtract 3) - 2)

5x - 3.50x = 120 - 97.50

1.50x = 22.5 x = 15

number of 10 oz. boxes are 15. answer is d.15</span>

6 0

2 years ago

Read 2 more answers

an employer will do a 50% match on your investment to a 401k retirement plan. If you decide to contribute a monthly amount of $2

faltersainse [42]

Answer:

The amount that should be in the account after 15 years is $95,321.85

Step-by-step explanation:

According to the given data, we have the following:

monthly amount of $220=R

interest rate is fixed at 2.05%. We require the monthly ineterest rate, hence monthly interest rate= 2.05%/12=0.1708%=0.0017

t=15years×12=180 months

In order to calculate how much should be in the account after 15 years, we would have to use the following formula:

Ap=<u>R(1-(1+i)∧-t)</u>

i

Ap=<u>220(1-(1+0.0017)∧-180)</u>

0.0017

Ap=<u>162,04</u>

0.0017

Ap=$95,321.85

The amount that should be in the account after 15 years is $95,321.85

<u />

6 0

2 years ago

Bob's bill for dinner at surefire steak house was $45. In addition he piad 5°\° of the bill in tax and he left a tip for 15°\° o

Anuta_ua [19.1K]

Answer:

$54

Step-by-step explanation:

First

45 times .05= 2.25

Second

45 plus 2.25

Third

45 times .15= 6.75

Fourth

45+2.25+6.75

5 0

2 years ago