You first have to turn 2x + 4y = -8 into y = mx + b before graphing. To do this, you subtract 2x from the left side of the equation because inverse operations. This will bring 2x to the right side. Now, you will have 4 = -2x + 8. However, you cannot have a number with y. Therefore, you must divide -8 and 2 by 4. You will get y = -1/2 x + 2 as your equation.
To graph -1/2x + 2, you must remember rise/run. Since it is NEGATIVE -1/2, your slope will go downwards (1 down, 2 across) starting from y = 2.
The X and Y angles created by lines intersection in the pictures are 18° and 54°.
Based on the picture, angle ∠MON is a right angle hence it has an 90° angle. We then know that the ∠MOA is 72°. Because angle ∠MOA lies within the angle ∠MON, hence we can write the following formula:
∠MON = ∠MOA +∠AON = 90°
∠MON = 72° + ∠AON = 90°
∠AON = 18° ... (i)
If we focus on the line CD being intersected by the line AB, hence we can conclude that the angles form by this intersection will follow these rules:
∠AOD = ∠BOC
∠AOC = ∠BOD
∠AOD + AOC = 180°
∠BOC + ∠BOD = 180°
Based on the picture, we know that:
∠BOC = x
∠AOC = ∠MOA + ∠MOC
∠AOC = 72° + y ...(ii)
∠AOD = ∠AON + ∠NOD
∠AOD = 18° +2x
∠BOC = 3x ... (iii)
Because we already know that ∠BOC = AOD, hence we could rewrite the formula into:
∠BOC = ∠AOD
3x = 18° + 2x
x = 18° ... (iv)
To find the value of y, we need to focus on angle ∠AOC. Based on the previous calculations and formulas, we know that:
∠AOC + ∠BOC = 180° ... (v)
Input equations (ii) and (iv) into (v)
∠AOC + ∠BOC = 180°
(72° + y) + 3x = 180°
72° + y + 3(18°) = 180°
126° + y = 180°
y = 54° ... (vi)
Learn more about the angles by lines intersection here: brainly.com/question/2077876?referrer=searchResults
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D = 20
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Answer:
Step-by-step explanation:
The perimeter of a polygon is equal to the sum of all the sides of the polygon. Quadrilateral PTOS consists of sides TP, SP, TO, and SO.
Since TO and SO are both radii of the circle, they must be equal. Thus, since TO is given as 10 cm, SO will also be 10 cm.
To find TP and SP, we can use the Pythagorean Theorem. Since they are tangents, they intersect the circle at a 
, creating right triangles 
and 
.
The Pythagorean Theorem states that the following is true for any right triangle:
, where 
is the hypotenuse, or the longest side, of the triangle
Thus, we have:
Since both TP and SP are tangents of the circle and extend to the same point P, they will be equal.
What we know:
Thus, the perimeter of the quadrilateral PTOS is equal to 
