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anastassius [24]

2 years ago

9

It takes 2 painters 12 hours to paint the interior of a house. Assuming the amount of time it takes to paint the house is invers

ely related to the number of painters, how many hours will it take 6 painters to paint the house?
A) 1 hour
B) 3 hours
C) 4 hours
D) 8 hours

2 answers:

I am Lyosha [343]2 years ago

4 0

Answer: Option 'C' is correct.

Step-by-step explanation:

Since we have given that

Number of painters to paint the interior of a house = 2

According to question, amount of time it takes to paint the house is inversely related to the number of painters.

So, Time is inversely related to number of painters.

If the number of painters to paint the house = 6

So, we get

$\frac{2}{6}=\frac{x}{12}\\\\2\times 12=6\times x\\\\24=6\times x\\\\\frac{24}{6}=x\\\\4\ hours=x$

Hence, it will take 4 hours to paint the house by 6 painters.

Therefore, Option 'C' is correct.

Sindrei [870]2 years ago

4 0

The answer to this question i 4

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What is the mean of the numbers 6 4 9 6 7 10 7

Paha777 [63]

Step-by-step explanation:

Mean =sum of values /number of values

total sum ↪️ 6+4+9+6+7+10+7

number of values is 7

= 49/7

<h2> = 7 is themean</h2>

6 0

2 years ago

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D is the trian

Alla [95]

Answer: mass (m) = 4 kg

center of mass coordinate: (15.75,4.5)

Step-by-step explanation: As a surface, a lamina has 2 dimensions (x,y) and a density function.

The region D is shown in the attachment.

From the image of the triangle, lamina is limited at x-axis: 0≤x≤2

At y-axis, it is limited by the lines formed between (0,0) and (2,1) and (2,1) and (0.3):

<u>Points (0,0) and (2,1):</u>

y = $\frac{1-0}{2-0}(x-0)$

y = $\frac{x}{2}$

<u>Points (2,1) and (0,3):</u>

y = $\frac{3-1}{0-2}(x-0) + 3$

y = -x + 3

Now, find total mass, which is given by the formula:

$m = \int\limits^a_b {\int\limits^a_b {\rho(x,y)} \, dA }$

Calculating for the limits above:

$m = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2(x+y)} \, dy \, dx }$

where a = -x+3

$m = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2} {(xy+\frac{y^{2}}{2} )} \, dx }$

$m = 2.\int\limits^2_0 {(-x^{2}-\frac{x^{2}}{2}+3x )} \, dx }$

$m = 2.\int\limits^2_0 {(\frac{-3x^{2}}{2}+3x)} \, dx }$

$m = 2.(\frac{-3.2^{2}}{2}+3.2-0)$

m = 2(-4+6)

m = 4

<u>Mass of the lamina that occupies region D is 4.</u>

<u />

Center of mass is the point of gravity of an object if it is in an uniform gravitational field. For the lamina, or any other 2 dimensional object, center of mass is calculated by:

$M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }$

$M_{y} = \int\limits^a_b {\int\limits^a_b {x.\rho(x,y)} \, dA }$

$M_{x}$ and $M_{y}$ are moments of the lamina about x-axis and y-axis, respectively.

Calculating moments:

For moment about x-axis:

$M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }$

$M_{x} = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2.y.(x+y)} \, dy\, dx }$

$M_{x} = 2\int\limits^2_0 {\int\limits^a_\frac{x}{2} {y.x+y^{2}} \, dy\, dx }$

$M_{x} = 2\int\limits^2_0 { ({\frac{y^{2}x}{2}+\frac{y^{3}}{3})}\, dx }$

$M_{x} = 2\int\limits^2_0 { ({\frac{x(-x+3)^{2}}{2}+\frac{(-x+3)^{3}}{3} -\frac{x^{3}}{8}-\frac{x^{3}}{24} )}\, dx }$

$M_{x} = 2.(\frac{-9.x^{2}}{4}+9x)$

$M_{x} = 2.(\frac{-9.2^{2}}{4}+9.2)$

$M_{x} = 18$

Now to find the x-coordinate:

x = $\frac{M_{y}}{m}$

x = $\frac{63}{4}$

x = 15.75

For moment about the y-axis:

$M_{y} = \int\limits^2_0 {\int\limits^a_\frac{x}{2} {2x.(x+y))} \, dy\,dx }$

$M_{y} = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2} {x^{2}+yx} \, dy\,dx }$

$M_{y} = 2.\int\limits^2_0 {y.x^{2}+x.{\frac{y^{2}}{2} } } \,dx }$

$M_{y} = 2.\int\limits^2_0 {x^{2}.(-x+3)+\frac{x.(-x+3)^{2}}{2} - {\frac{x^{3}}{2}-\frac{x^{3}}{8} } } \,dx }$

$M_{y} = 2.\int\limits^2_0 {\frac{-9x^3}{8}+\frac{9x}{2} } \,dx }$

$M_{y} = 2.({\frac{-9x^4}{32}+9x^{2})$

$M_{y} = 2.({\frac{-9.2^4}{32}+9.2^{2}-0)$

$M{y} =$ 63

To find y-coordinate:

y = $\frac{M_{x}}{m}$

y = $\frac{18}{4}$

y = 4.5

<u>Center mass coordinates for the lamina are (15.75,4.5)</u>

3 0

3 years ago

Find the distance and midpoints of the following points:

timofeeve [1]

1.) (2, 4) and (10, 8)
Distance: 4√5
Midpoint: (6, 6)

2.) (3, 8) and (7, 3)
Distance: √41
Midpoint: (5, 11/2)

3.) (4, 9) and (9, 5)
Distance: √41
Midpoint: (13/2, 7)

4 0

3 years ago

Can someone plz help me. How can you find the inequalities of 11/15 and 5/7. Next 5/9 and 7/13. Next 11/15 and 5/7. Lastly 5/9 a

LekaFEV [45]

To make this a little clearer, let's give the pairs of inequalities the same denominator:

<span>Question 1:
</span> $\frac{11}{15}$ ? $\frac{5}{7}$
First, apply the common denominator to the first fraction:
$(\frac{11}{15})7 \\ \frac{11}{15} * \frac{7}{7} \\ \frac{11*7}{15*7} \\ \frac{77}{105}$
Do the same for the second: $15( \frac{5}{7}) \\ \frac{5}{7}* \frac{15}{15} \\ \frac{5*15}{7*15} \\ \frac{75}{105}$
Nest, compare the two fractions:
$\frac{77}{105} \ \textgreater \ \frac{75}{105}$
Therefore:
$\frac{11}{15}$ > $\frac{5}{7}$
<span>
Question Two:</span>
$\frac{5}{9}$ ? $\frac{7}{13}$
Apply the common denominator to fraction one:
$13( \frac{5}{9}) \\ \frac{5}{9} * \frac{13}{13} \\ \frac{5*13}{9*13} \\ \frac{65}{117}$
Fraction two:
$9(\frac{7}{13}) \\ \frac{7}{13} * \frac{9}{9} \\ \frac{7*9}{13*9} \\ \frac{63}{117}$
Evaluate:
$\frac{65}{117}$ > $\frac{63}{117}$
Therefore:
<span> $\frac{5}{9}$ > $\frac{7}{13}$
</span>
Hope this helps!

5 0

3 years ago

Mark buys a wooden board that is 7 1/2 feet long. The cost of the board is $0.50 per foot, including tax. What is the total cost

artcher [175]

Answer:

Total cost of Mark's board is $3.75.

Step-by-step explanation:

We are given the following in the question:

Length of wooden board =

$=7\dfrac{1}{2}\text{ foot}$

Unit cost of board = $0.50 per foot

Total cost of board =

$=\text{Length of board}\times \text{Unit cost of board}\\\\=7\dfrac{1}{2}\times 0.50\\\\=\dfrac{15}{2}\times \dfrac{1}{2}\\\\=\dfrac{15}{4} = 3.75\$$

Thus, total cost of Mark's board is $3.75.

7 0

2 years ago