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zhannawk [14.2K]
2 years ago
10

What is the value of the expression below 1/4 = 8? (5.NF.7)

Mathematics
1 answer:
alexgriva [62]2 years ago
8 0
The answer would be 1/32
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find the value of c that makes each trinomial a perfect square. then write the trinomial as a perfect square. the equation , x^2
mr Goodwill [35]
Therefore, the value of C should be (11/2)^2

<span>To square a fraction, take the square of the numerator and put it over the square of the denominator. So since 11^2=121 and 2^2=4, (11/2)^2
(11)^2/(2)^2=121/4
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
3 0
3 years ago
Which statements are true regarding the area of circle
Pavlova-9 [17]

<u>Question:</u>

Which statements are true regarding the area of circle D? Select two options.

The area of the circle depends on the square of the radius.

The area of circle D is 36Pi centimeters squared.

The area of circle D is 324Pi centimetres squared

The area of the circle depends on the square of pi.

The area of the circle depends on the central angle.

<u>Answer:</u>

<u>The following statements are true regarding the area of circle D: </u>

  • The area of the circle depends on the square of the radius
  • The area of circle D is 324Pi centimeters

<u>Step-By-Step Explanation:</u>

As we know, the formula for area of the circle is given by,

             A=\pi r^{2}

Where, r = radius of the circle. In figure, CD = radius = 18 cm

So, the circle area can be measured as

            A=\pi(18)^{2}=324 \pi \mathrm{cm}^{2}

From this, concluding that the circle area depends on radius square and its value is 324 \pi \mathrm{cm}^{2}. Hence, the statement given in option A and C are correct (reveals that option B is incorrect).  

As the Pi is constant, the circle radius cannot depend on Pi value and Option D also incorrect. The area of sector of circle only depends on ratio of central angle to entire circle (So option E also incorrect).

6 0
3 years ago
Read 2 more answers
Alvin is 7 years older than Elga. The sum of their ages is 97. What is Elga's age?
LenaWriter [7]

Elga's age = x

Alvin's age = x + 7

x + x + 7 = 97.

Simplify the left side of the equation

2x + 7 = 97.

Subtract 7 from each side

2x = 90.

Divide each side by 2

x = 45

Since x is equal to Elga's age, Elga is 45 years old

4 0
3 years ago
What is the equation of the line that passes through (-3, -1) and has a slope of 3/5 ? Put your answer in slope-intercept form.
Verizon [17]

Slope-intercept form: y = mx + b

m = the slope

b = y-intercept.

In this problem,

m = 3/5

b = ?

So far y = 3/5x + b

Let's plug the point (-3, -1) into our slope equation.

-1 = 3/5(-3) + b

Simplify the right side.

-1 = -9/5 + b

Add 9/5 to both sides.

4/5 = b

The equation is: y = 3/5x + 4/5

Answer choice A is correct.

6 0
3 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
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