This is solvable only if e b is the initial side and b d and b c lies on opposite side of each other and lies on a line i.e c,b,d are Collinear.
∠ebd and ∠ebc will form a linear pair.The meaning of linear pair is that angles forming on one side of a straight line through a common vertex which are adjacent is 180°.
i.e
∠ ebd + ∠ebc = 180°
4 x- 8 + 5x + 20= 180°
adding like terms
⇒ 9 x +12 =180°
⇒ 9 x = 180° - 12
⇒ 9 x = 168°
⇒ x =( 168/9)°=(56/3)°
now m□ebc =5 x +20
= 5 × 56/3 + 20
= 280/3 + 20
=340/3
m□ebc=( 340/3)°
So, solution set is x =(56/3)° and m□ebc =(340/3)°
A slope indicates the steepness of a line while the intercept points the location where it intercepts its axis. The linear relationship between can be defined using the intercept and the slope. Both concepts are used to estimate the average range of change. Since we are trying to add a peak current value of 0.38 which is lesser than the average, the intercept of the graph would therefore decrease and the slope increase.
We know that If a system has at least one solution, it is said to be consistent. When you graph the equations, both equations represent the same line so the system has an infinite number of solutions
If a consistent system has an infinite number of solutions, it is dependent. <span> therefore the system is </span>consistent, dependent and <span>equivalent </span><span> the answer is </span>equivalent
The simplest fraction for is . Write the upper bound as a fraction with the same denominator:
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Hence the range for would be:
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If the denominator of is also , then the range for its numerator (call it ) would be . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than .
To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)
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At this point, the difference between the numerators is now . That allows a number ( in this case) to fit between the bounds. However, can't be written as finite decimals.
Try multiplying the numerator and the denominator by a different number.
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It is important to note that some expressions for can be simplified. For example, because of the common factor .
The circle is inscribed in triangle PRT. A circle is inscribed in triangle P R T. Points Q, S, and U of the circle are on the sides of the triangle. Point Q is on side P R, point S is on side R T, and point U is on side P T. The length of R S is 5, the length of P U is 8, and the length of U T is 6. Which statements about the figure are true?
Answer:
(B)TU ≅ TS
(D)The length of line segment PR is 13 units.
Step-by-step explanation:
The diagram of the question is drawn for more understanding,
The theorem applied to this problem is that of tangents. All tangents drawn to a circle from the same point are equal.