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Kobotan [32]
3 years ago
11

A rectangle is placed around a semicircle as shown below. The length of the rectangle is 14mm. Find the area of the shaded regio

n. Use the value 3.14 for pi , and do not round your answer. Be sure to include the correct unit in your answer.

Mathematics
1 answer:
Damm [24]3 years ago
7 0

Image is missing, so i have attached it

Answer:

21.07 mm²

Step-by-step explanation:

We will use the Formulas:

Area of semicircle = πr²/2

Area_rectangle = Length x width = Lw

We are given:

Length of rectangle = 14 mm

From the attached image, the diameter of the semicircle is equal to the length of the rectangle.

Thus;

diameter of semicircle = length of rectangle

Therefore,

diameter of semicircle = 14 mm

Also, from the attached image, the radius of the semicircle is equal to the width of the rectangle.

Thus;

diameter of semicircle/2 = radius of semicircle

Hence;

radius of semicircle = 14/2 = 7mm = width of rectangle

Thus;

Area of rectangle = L*w = 14 × 7 = 98 mm²

Area of semicircle = πr²/2 = π×7²/2 = 76.93 mm²

To get the area of the shaded region, we will subtract area of the semi-circle from the area of the rectangle.

Area of shaded region = Area_rectangle - Area_semicircle = 98 - 76.93 = 21.07 mm²

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The half life of silicone-32 is 710 years. If 30 grams is present now, how much will be present in 300 years?
Irina18 [472]

Answer:

22.38 g of silicone-32 will be present in 300 years.

Step-by-step explanation:

A radioactive half-life refers to the amount of time it takes for half of the original isotope to decay and its given by

                                           N(t)=N_0(\frac{1}{2})^{\frac{t}{t_{1/2}}

where,

N(t) = quantity of the substance remaining

N_0 = initial quantity of the substance

t = time elapsed

t_{1/2} = half life of the substance

From the information given we know:

  • The initial quantity of silicone-32 is 30 g.
  • The time elapsed is 300 years.
  • The half life of silicone-32 is 710 years.

So, to find the quantity of silicone-32 remaining we apply the above equation

N(t)=30\left(\frac{1}{2}\right)^{\frac{300}{710}}=30\left(\frac{1}{2}\right)^{\frac{30}{71}}\approx22.38 \:g

22.38 g of silicone-32 will be present in 300 years.

7 0
3 years ago
Help quick ❗️❗️ Which sequence of transformations will change did your PQRS to figure P’Q’R’S’?
babymother [125]

Answer:

Counterclockwise rotation about the origin by 90° followed by reflection over the Y-axis ⇒ answer (D)

Step-by-step explanation:

* Lets revise the reflection and the rotation of a point

- If point (x , y) reflected across the x-axis

∴ Its image is (x , -y)

- If point (x , y) reflected across the y-axis

∴ Its image is (-x , y)

- If point (x , y) rotated about the origin by angle 90° counter clockwise

∴ Its image is (-y , x)

- If point (x , y) rotated about the origin by angle 90° clockwise

∴ Its image is (y , -x)

- If point (x , y) rotated about the origin by angle 180°

∴ Its image is (-x , -y)

* There is no difference between rotating 180° clockwise or  

anti-clockwise around the origin

- In our problem

# The vertices of the original figure are:

  P(1 , -3) , Q (3 , -2) , R (3 ,-3) , S (2 , -4)

# The vertices of the image are:

  P' (-3 , 1) , Q' (-2 , 3) , R' (-3 , 3) , S' (-4 , 2)

∵ x and y are switched

∴ The figure is rotated about origin by 90°

∵ The signs of the x-coordinates and the y-coordinates

   didn't change

∴ The rotation is followed by reflection, to know about which

   axis we must to find the images after the rotation

∵ P (1 , -3) ⇒ after rotation 90° counter clockwise is (3 , 1)

∵ Q (3 , -2) ⇒ after rotation 90° counter clockwise is (2 , 3)

∵ R (3 , -3) ⇒ after rotation 90° counter clockwise is (3 , 3)

∵ S (2 , -4) ⇒ after rotation 90° counter clockwise is (4 , 2)

* The images are P' (-3 , 1) , Q' (-2 , 3) , R' (-3 , 3) , S' (-4 , 2)

- The sign of x-coordinates of all of them changed

∴ The rotation followed by reflection over the y-axis

* The answer is ⇒ Counterclockwise rotation about the origin

  by 90° followed by reflection over the Y-axis

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Answer:

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Step-by-step explanation:

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vlabodo [156]

Given equation is y=2x+7 and \frac{1}{2}y=x+\frac{7}{2}

Let's simplify the 2nd equation \frac{1}{2}y=x+\frac{7}{2} before we can start graph so that calculation will be easy

\frac{1}{2}y=x+\frac{7}{2}

multiply both sides by 2 to cancel out fractions

2*\frac{1}{2}y=2*x+2*\frac{7}{2}

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