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Tresset [83]
3 years ago
7

The following angles in a convex quadrilateral Angle A= 56.1 degrees Angle B= 99.4 degrees Angle C= 78.2 degrees What is the mea

sure of the missing angle?
Mathematics
1 answer:
Contact [7]3 years ago
7 0

Hello,

The sum of angles in a quadrilateral is 360°.

56,1° + 99,4° + 78,2° + x = 360°

233,7° + x = 360°

x = 360° - 233,7° = 126,3°

Bye :)

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A tennis ball can in the shape of a right circular cylinder holds four tennis balls snugly. If the radius of a tennis ball is 3.
Volgvan

Answer:

  about 33.3%

Step-by-step explanation:

The fraction of space not occupied will be the same as the unoccupied fraction of a cylinder of the same height and diameter as a sphere.

<h3>Volume of a cylinder</h3>

The volume of a cylinder of diameter d and height d will be ...

  V = πr²h = π(d/2)²(d) = (π/4)d³

<h3>Volume of a sphere</h3>

The volume of a sphere of diameter d is ...

  V = 4/3πr³ = 4/3π(d/2)³ = (π/6)d³

<h3>Unoccupied space</h3>

The fraction of space that is occupied is ...

  occupied space = (sphere volume)/(cylinder volume)

  occupied space = ((π/6)d³) / ((π/4)d³) = 4/6 = 2/3

The fraction of unoccupied space is ...

  unoccupied fraction = 1 -2/3 = 1/3

  unoccupied fraction ≈ 33.3%

3 0
2 years ago
Read 2 more answers
The point (-4, -1), is the bottom of a triangle. Which point would it map to if the triangle was translated right 5 units and re
lyudmila [28]

Answer:

Step-by-step explanation:

(x,y)→(x+5,y)

(-4,-1)→(-4+5,-1)=(1,-1)

reflection about x-axis

(x,y)→(x,-y)

(1,-1)→(1,1)

8 0
3 years ago
Read 2 more answers
A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot
ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box  is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = (x\times x)

                               =x^2

The volume of the box  is = area of the base × height

                                           =x^2h

Therefore,

x^2h=272

\Rightarrow h=\frac{272}{x^2}

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

=20x^2 cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

=10x^2 cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

=4xh\times 1.5 cents

=6xh cents

Total cost = (20x^2+10x^2+6xh)

                =30x^2+6xh

Let

C=30x^2+6xh

Putting the value of h

C=30x^2+6x\times \frac{272}{x^2}

\Rightarrow C=30x^2+\frac{1632}{x}

Differentiating with respect to x

C'=60x-\frac{1632}{x^2}

Again differentiating with respect to x

C''=60+\frac{3264}{x^3}

Now set C'=0

60x-\frac{1632}{x^2}=0

\Rightarrow 60x=\frac{1632}{x^2}

\Rightarrow x^3=\frac{1632}{60}

\Rightarrow x\approx 3

Now C''|_{x=3}=60+\frac{3264}{3^3}>0

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is =\frac{272}{3^2}

                                          =30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

7 0
3 years ago
In fig if x+y=w+z then prove that AOB is a line.​
frez [133]

Answer:

As Given, x+y=w+z

To Prove: AOB is a line or x+y=180

∘

(linear pair.)

According to the question,

x+y+w+z=360

∘

∣ Angles around a point.

(x+y)+(w+z)=360

∘

(x+y)+(x+y)=360

∘

∣ Given x+y=w+z

2(x+y)=360

∘

(x+y)=180

∘

Hence, x+y makes a linear pair.

Therefore, AOB is a straight line

4 0
3 years ago
Help please!! The measures of the angles of a triangle are shown in the figure below. solve for x
tatiyna

9514 1404 393

Answer:

  x = 33

Step-by-step explanation:

The angles are complementary.

  x° = 90° -57°

  x = 33

3 0
3 years ago
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