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2 years ago

9

A rectangular blankets perimeter is 210 inches. If the long sides of the blanket measure 60 inches what is the length of the sho

rter side of the blanket

1 answer:

hram777 [196]2 years ago

3 0

I believe the answer is 45

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The Colossus Ferris wheel debuted at the 1984 New Orleans World’s Fair. The ride is 180 ft tall, and passengers board the ride a

Makovka662 [10]

Answer:

$T=\frac{2}{3}$

Interpretation:

It means that Passenger will reach to same height after each 2/3 minutes

Step-by-step explanation:

We are given height function as

$h=82.5sin(3\pi(t+0.5))+97.5$

where

h is the height above the ground

t is the time

we can compare it with standard equation

$y=Asin(B(t+C))+D$

Period formula is

$T=\frac{2\pi}{B}$

now, we can compare and find B

$B=3\pi$

we can find period

$T=\frac{2\pi}{3\pi}$

$T=\frac{2}{3}$

Interpretation:

It means that Passenger will reach to same height after each 2/3 minutes

5 0

2 years ago

Evaluate e − 1 2 f e− 2 1 fe, minus, start fraction, 1, divided by, 2, end fraction, f when e = 15 e=15e, equals, 15 and f = 2

ivolga24 [154]

The answer 456 fe
Because it the benefit is doing today

7 1

2 years ago

Read 3 more answers

A dartboard has 8 sections of equal area. The letters represent colors red (R), yellow (Y), blue (B), and green (G). Complete th

Alla [95]

Since the sum of all probabilities of all all elementary events will always be equal to 1. Furthermore, the probabilities of all mutually exclusive set of events that is part of the entire sample space will always be total of 1.

So in the problem, the answer is 1/8.
1/8 for red + 3/8 for green + 3/8 for yellow + 1/8 for blue = 8/8 or 1.

4 0

1 year 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

Ed is 7 years older than Ted. Ed’s age is also 3/4 times Ted’s age. How old are Ed and Ted?

svetlana [45]

Let Ted be x.

Ed is 7 years older = x + 7

Ed = (3/4)Ted

(x + 7) = (3/4)x

x + 7 = 3x/4

x - 3x/4 = -7

x/4 = -7

x = -28, Ted = -28 years.

(x + 7) = -28 + 7 = -21, Ed = -21 years

Goodness. We had negative numbers for the ages, well does that make sense? No it doesn't.

Our answer is correct. But the sense in the question is lacking. The question has been wrongly set.

<span>We might assume negative ages to mean before they came into the world, before birth! </span>

7 0

2 years ago