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sesenic [268]
2 years ago
13

A cylindrical container that contains three tennis balls that are 2. 7 inches in diameter is lying on its side and passes throug

h an x-ray scanner that scans from the top only. What outline will appear on the x-ray screen, and what are the dimensions? cylindrical container with three spheres side by side so that the spheres are laying horizontally A circle with a diameter of 2. 7 in A circle with a diameter of is 5. 4 in A rectangle that is 5. 4 in x 8. 1 in A rectangle that is 2. 7 in x 8. 1 in.
Mathematics
1 answer:
kirza4 [7]2 years ago
5 0

The correct option is c, which is A rectangle that is 5.4 in x 16.2 in.

<h3>Rectangle</h3>

A rectangle is a quadrilateral with opposite sides parallel and equal to each other. Also, all the angles of the rectangle are equal to 90°.

Given

A cylindrical container that contains three tennis balls that are 2. 7 inches in diameter is lying on its side and passes through an x-ray scanner that scans from the top only.

As given, the cylinder is lying on its side, thus for the x-ray machine, the side will be the top of the cylinder. therefore, the cylinder will look like a rectangle, like the way it is shown below.

The diameter of 3 balls will be the length of the rectangle while the width will be the diameter of 1 ball.

Hence, the correct option is c, which is A rectangle that is 5.4 in x 16.2 in.

Learn more about Cylinder:

brainly.com/question/3692256

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X+ y +z = 37
telo118 [61]

Answer:

C

Step-by-step explanation:

I'm not sure of my answer i hope it helps

4 0
3 years ago
I dont know man help please
Ratling [72]

Answer:

m∠1 = 63°

m∠2 = 49°

m∠3 = 87°

m∠4 = 44°

Step-by-step explanation:

From the given figure

∵ ∠2 and 49° are vertically opposite angles

∵ The vertical opposite angles are equal in measures

∴ m∠2 = 49°

∵ The sum of the interior angles of a Δ is 180°

∵ m∠1, m∠2, and 68° are interior angles of a Δ

∴ m∠1 + m∠2 + 68° = 180°

∵ m∠2 = 49°

∴ m∠1 + 49° + 68° = 180°

→ Add the like terms in the left side

∴ m∠1 + 117 = 180

→ Subtract 180 from both sides

∴ m∠1 = 63°

∵ ∠3 and 93° formed a pair of linear angles

∵ The sum of the measures of the linear angles is 180°

∴ m∠3 + 93° = 180°

→ Subtract 93 from both sides

∴ m∠3 = 87°

∵ 93° is an exterior angle of the triangle

∵ The measure of the exterior angle of a Δ at one vertex equals the sum

   of the measures of the opposite interior angles to this vertex

∵ ∠4 and 49° are the opposite interior angles to 93°

∴ 49° + m∠4 = 93°

→ Subtract 49 from both sides

∴ m∠4 = 44°

8 0
3 years ago
Find the slope between the following two points: (3, 1), (5, 4).
gulaghasi [49]

Answer:

3/2 or 1.5

Step-by-step explanation:

slope =  \frac{4 - 1}{5 - 3}  \\  \\  =  \frac{3}{2}  \\  \\  = 1.5

4 0
2 years ago
Read 2 more answers
Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi
kompoz [17]

Answer:

The general solution is

y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))

     + \frac{x^2}{2}

Step-by-step explanation:

Step :1:-

Given differential equation  y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

(D^4 -2D^3+D^2)y = e^x+1

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

                         m^4 -2m^3+m^2 = 0

                         m^2(m^2 -2m+1) = 0

(m^2 -2m+1) this is the expansion of (a-b)^2

                        m^2 =0 and (m-1)^2 =0

The roots are m=0,0 and m =1,1

complementary function is y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x

<u>Step 2</u>:-

The particular equation is    \frac{1}{f(D)} Q

P.I = \frac{1}{D^2(D-1)^2} e^x+1

P.I = \frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}

P.I = I_{1} +I_{2}

\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}

\frac{1}{D} means integration

\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx

applying in integration u v formula

\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx  } )\, dx

I_{1} = \frac{1}{D^2(D-1)^2} e^x

\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))

\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))

I_{2}= \frac{1}{D^2(D-1)^2}e^{0x}

\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x

again integration  \frac{1}{D} x = \frac{x^2}{2!}

The general solution is y = y_{C} +y_{P}

         y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))

      + \frac{x^2}{2!}

3 0
3 years ago
If a rectangle A has sides are three times the length of those rectangle B, how do the areas of the two rectangles compare?
Radda [10]

Answer:

The area of A is nine times the area of B.

Step-by-step explanation:

4 0
2 years ago
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