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Flura [38]
2 years ago
12

Scott Peter’s bank granted him a single-payment loan of $3,250 to pay a repair bill. He agreed to repay the loan in 31 days at a

n exact interest rate of 11.75%. What is the maturity value of the loan? a $3,300.90 b $3,211.09 c $3,282.43 d $3,554.66
Mathematics
1 answer:
kifflom [539]2 years ago
5 0
Im pretty sure the answer is C.
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Ramon has 6.75 kg coffee. he wants 0.25 kg bags. how many bags does he need?
Hatshy [7]

Answer:

Ramon would need 27 bags

Step-by-step explanation:

This is because if you divide 6.75 by 0.25 you would get 27

5 0
2 years ago
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 vectora=(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
Which descriptions from the list below accurately describe the relationship between QRS and TUV
Nata [24]
Answer is A. similar

hope it helps
8 0
2 years ago
Read 2 more answers
In circle M shown, chords GH and EF intersect at K such that EK  5 and FK  6 . If GK  3 , then which of the following is the
myrzilka [38]

The length of the GH segment is 13  

Step-by-step explanation:

For solving this problem we need to remember some of the circle corollaries-

When two-chord intersects each other, the product of the chord segments are equal

The above corollary can be easily understood by looking at a diagram attached below-

In the figure, EF and GH are two chords intersecting at K

Thus, EK*KF= GK*KH

Values of the EK, KF, GK are given as 5, 6 and 3 respectively

Substituting the values we get

5*6=3*KH

KH= 10

We know that GH= GK+KH

Thus GH= 3+10= 13

6 0
2 years ago
A solid lies between two parallel planes 22 feet apart and has a volume of 4343 cubic feet. What is the average area of cross-se
ella [17]

Answer:

Step-by-step explanation:

So, formula to be used is:

Average area of cross section is = volume/distance between plates.

Volume = 4343/22 = 197.4 square feet.

8 0
2 years ago
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