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1 year ago

11

Bob and Ted are in the plumbing business. Bob can do the plumbing hook-up for a new house in 4 hours; Ted does the same job in 6

hours. If they work together, how long will they take to hook up the plumbing in one house? 2.1 hours 2.4 hours 5 hours

2 answers:

sladkih [1.3K]1 year ago

8 0

Answer:

2.4 hours

Step-by-step explanation:

Stels [109]1 year ago

5 0

Where 1/4 represents Bob and 1/6 represents Ted...
1/4x + 1/6x = 1
Multiply the two so there is no fraction by the GCF of 12.
3x + 2x = 12
Add 3x and 2x.
5x = 12
Divide by 5.
x = 12
Plug this value in.
With that, the answer is 2 hours and 24 minutes with x = 12 meaning 144 minutes.

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Answer:

d and f 2.a and c 3.b and e

Step-by-step explanation:

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1 year ago

In triangle ABC segment DE is parallel to the side AC. (The endpoints of segment DE lie on the sides AB and BC respectively).

ivanzaharov [21]

Answer:

6 dm

Step-by-step explanation:

Triangle DBE is similar to triangle ABC, so their side lengths are proportional.

DE/AC = DB/AB

The length of DB can be found from ...

DB +AD = AB

DB = AB -AD = (15 -10) dm = 5 dm

So, we can fill in the proportion:

DE/(18 dm) = (5 dm)/(15 dm)

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1 year ago

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

The figure below shows the dimensions of a city park in feet. Shannon thinks the area of city park is 73,084 square feet. Do you

Andreyy89

Answer:

We need to solve for the 4th side

4th side base = 75.5 -60.5 = 10 feet

4th side height = 16

4th side LENGTH^2 = 10^2 + 16^2

4th side = sq root (356) = 18.8679622641

Trapezoid Area = [(sum of the bases) / 2 ] * height

Trapezoid Area = [(136)/2] * 16

Trapezoid Area = 1,088 square feet, which is MUCH smaller

than 73,084

Step-by-step explanation:

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1 year ago

Which equation represents a line that passes through (5, 1) and has a slope of StartFraction one-half EndFraction?

saul85 [17]

Answer:

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Step-by-step explanation:

we know that

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=\frac{1}{2}$

$(5,1)$

substitute

$y-1=\frac{1}{2}(x-5)$

5 0

1 year ago

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