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3 years ago

Angle ACB is an inscribed angle of circle P. Find x if angle ACB measures x° and arc ACB measures (x + 45)° .

1 answer:

8 0

According to the theorem of circles, angle ACB is equal to 1/2 of the explementary angle of angle APB. The explementary angle then is 360 - (x+45) or 315 - x where x is angle ACB. Hence x = 0.5 * (315-x). x is equal to 105 degrees.

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4 years ago

Statistical studies are not concerned with understanding the sources of variability in data, only with describing the variabili

BaLLatris [955]

Answer:

Step-by-step explanation:

it's B

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3 years ago

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Which has the greater surface area? A 6-centimeter cube or a rectangular prism with a 3 centimeter square base and a height of 1

bazaltina [42]

Answer:

The prism

Step-by-step explanation:

The surface area of a cube is:

$A=6L^2$

where

L is the length of the side

For the cube in this problem,

L = 6 cm

So the surface area of the cube is

$A=6\cdot 6^2 = 36 cm^2$

The surface area of a rectangular prism with squared base is:

$A=2L^2+4Lh$

where

L is the length of the base

h is the height of the prism

For the prism in this problem,

L = 3 cm

h = 12 cm

Therefore,

$A=2\cdot 3^2 + 4(3)(12)=162 cm^2$

So, the prism has a greater surface area.

3 0

3 years ago

Helpppp pleaseeeeeeeeeeeeeee

Vesna [10]

Answer:

Find below the calculations of the two areas, each with two methods. The results are:

Upper triangle:

$Area=5000\sqrt{3}units^2$

Lower triangle:

$Area=14,530m^2$

Explanation:

A) Method 1

When you are not given the height, but you are given two sides and the included angle between the two sides, you can use this formula:

$Area=side_1\times side_2\times sin(\alpha)$

Where, $\alpha$ is the measure of the included angle.

1. Upper triangle:

$side_1=200units\\ \\ side_2=100units\\ \\ \alpha =60\º\\ \\ Area=200units\times 100units\times sin(60\º)/2\\ \\ Area=5000\sqrt{3}units^2$

2. Lower triangle:

$side_1=231m\\ \\ side_2=150m\\ \\ \alpha =123\º\\ \\ Area=231m\times 150m\times sin(123\º)/2\\ \\ Area=14,529.96m^2\approx14,530m^2$

B) Method 2

You can find the height of the triangle using trigonometric properties, and then use the very well known formula:

$Area=(1/2)\times base\times height$

Use it for both triangles.

3. Upper triangle:

The trigonometric ratio that you can use is:

$sine(\alpha)=opposite\text{ }leg/hypotenuse$

Notice the height is the opposite leg to the angle of 60º, and the side that measures 100 units is the hypotenuse of that right triangle. Then:

$sin(60\º)=height/100units\\ \\ height=sin(60\º)\times100units\\ \\ height=50\sqrt{3}units$

$Area=(1/2)\times base\times height=(1/2)\times 200units\times 50\sqrt{3}units=5,000\sqrt{3}units^2$

3. Lower triangle:

$sin(180\º-123\º)=height/231m\\ \\ height=sin(57\º)\times 231m\\ \\ height=193.7329m^2$

$Area=(1/2)\times base\times height=(1/2)\times 150m\times 193.7329m^2\\\\ Area=14,529.96m^2\approx 14,530m^2$

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3 years ago

Which of the following is not a way to represent the solution of the inequality 2(x − 1) less than or equal to 10? (1 point)

mars1129 [50]

The answer is D because A and B mean the exact same thing and if A and B are graphed the result is C so the only logical answer left is D

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3 years ago

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