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2 years ago

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One year ago, Lindsey deposited $250 into a savings account. Her balance is now $253. Two years ago, Jenn deposited $250 into a

savings account. Her balance is now $257.50. Which account has the greater simple interest rate? Explain.

2 answers:

VARVARA [1.3K]2 years ago

8 0

275.50 it is greater than the other

Klio2033 [76]2 years ago

8 0

Answer:

275.50 it is greater than the other

Step-by-step explanation:

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I NEED HELP PLEASE! Jenny's birthday cake is circular and has a 30 cm radius. Her slice creates an arc of with a central angle o

dybincka [34]

Ok, so if the angle is 120 degrees, its one third of the full cake (which is 360 degrees). So the area of the whole circle (worked out with area = pi * r^2 where r is 30) gives 900*pi, and so one third of that (because her slice is one third) is 300pi.

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2 years ago

Classify the solution set of each equation below as one of the following types of surfaces: A. parabolic cylinder, B. ellipsoid,

taurus [48]

Answer:

the complete question is found in the attachment

1) D. hyperbolic paraboloid

2)C. elliptic paraboloid

3)E. cone

4)F. hyperboloid.

Step-by-step explanation:

The complete explanation is found in the attachment

3 0

1 year ago

If Chung's go-kart can travel 48 miles on 1 gallon of gas, how far can his go-kart travel on 1.15 gallons of gas?

Nataliya [291]

So to get the answer you multiply 48 by 1.15 and you get 55.2
Chung's go-kart can travel 55.2 miles on 1.15 gallons of gas.

3 0

2 years ago

Read 2 more answers

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

Bas_tet [7]

Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

$\bar{x}=\frac{21}{168}*40=5$

Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

3 0

2 years ago

A barber shop produces 96 haircuts a day. each barber in the shop works 8 hours per day and produces the same number of haircuts

Finger [1]

We know that
A barber shop produces 96 haircuts a day----------> <span>8 hours per day
so
96/8--------> 12 haircuts per hour

if </span><span>if the shop's productivity is 2 haircuts per hour of labor
then
12/2---------> 6 </span><span>barbers

the answer is
6 </span><span>barbers</span>

7 0

2 years ago