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kodGreya [7K]
3 years ago
5

Hi! I just need someone to check if this is correct I also have to show the work so if you can that is much appreciated! The per

imeter of a square is one yard. What is the area of the square in square inches? My answer is 36
Mathematics
1 answer:
Anestetic [448]3 years ago
7 0

Perimeter (P) of a square = 4s ; where s is the length of a side

1 yd = 4s

\frac{1}{4} yd = s

Area (A) of a square = s²

= (\frac{1 yd}{4})²

Now, convert from yard to foot to inch:

(\frac{1 yd}{4})² x (\frac{3 ft}{1 yd})² x (\frac{12 in}{1 ft})²

= (3 x 3 x 12 x 12) ÷ (4 x 4)

= 81 in²

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A number divided by five is equal to four . what is the number ?
aliya0001 [1]

Answer:

20

Step-by-step explanation:

5 times 4 is 20 so therefore 20 divided by 4 is 5 or any way around

7 0
3 years ago
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Simplify the following expression: (9x^2+2x−5)+(6x^2−5x+4). Please explain how you found your answer
sergeinik [125]
Hi there!

Simplify th' Eqn.

(9x² + 2x - 5) + (6x² - 5x + 4)

⇒ 9x² + 2x - 5 + 6x² - 5x + 4

⇒9x² + 6x² + 2x - 5x - 5 + 4

⇒ 15x² - 3x - 1

Hence,
The simplified form is = 15x² - 3x - 1

~ Hope it helps!
5 0
3 years ago
A road is 8/9 of a mile long. A crew needs to repave 2/5 of the road. How long is the section that needs to be repaved?
Rama09 [41]

Answer:

16/45

Step-by-step explanation:

Explanation is in photo

8 0
2 years ago
Find the value of y in the equation attached to this question
dlinn [17]

Answer:

y = 4

Step-by-step explanation:

1) Simplify 3^5 to 243.

{9}^{y}=\frac{243\times {9}^{6}}{{27}^{3}}

2) Simplify {9}^{6} to 531441.

{9}^{y}=\frac{243\times 531441}{{27}^{3}}

3)  Simplify  243\times 531441 to  129140163.

{9}^{y}=\frac{129140163}{{27}^{3}}

4) Simplify  {27}^{3}  to  19683.

{9}^{y}=\frac{129140163}{19683}

5) Simplify  \frac{129140163}{19683} to 6561.

{9}^{y}=6561

6) Convert both sides to the same base.

9^y=9^4

7)  Cancel the base of 9 on both sides.

y=4

Cheers,

ROR

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 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
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