Consider the circle with center X, as shown in the figure.
Draw the diameter of the circle which is parallel to cherd AB, as shown in the figure.
Since the diameter and AB are parallel, then the line segment XC which bisects AB at C, will be perpendicular to AB.
SO triangle XCB is a right triangle. Thus the length of CX, by the Pythagorean theorem is
units.
Answer: 8 units
Answer:
I think its D
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given:
°
From the triangle, using the theorem that center angle by an arc is twice the angle it subtend at the circumference.
Also, the diameter of the circle is BD. As per the theorem that says that angle subtended by the diameter at the circumference is always 90°,
From the Δ BCD, which is a right angled triangle,
Now, using the theorem that angle between the tangent and a chord is equal to the angle subtended by the same chord at the circumference.
Here, chords CD and BC subtend angles 40 and 50 at the circumference as shown in the diagram by angles 
and EF is a tangent to the circle at point C.
Therefore, 
Again, using the same theorem as above,
Hence, all the angles are as follows:
Answer:
Step-by-step explanation:
- 13t - 3t = (13 - 3)t = 10t
- 3t - 13t = (3 - 13)t = -10t
- 10t ≠ -10t
They are not equivalent
<u>When t = 2</u>
- 13t - 3t = 13*2 - 3*2 = 26 - 6 = 20
- 3t - 13t = 3*2 - 13*2 = 6 - 26 = -20
Answer:
Step-by-step explanation:
Answer:
a) y-8 = (y₀-8) , b) 2y -5 = (2y₀-5)
Explanation:
To solve these equations the method of direct integration is the easiest.
a) the given equation is
dy / dt = and -8
dy / y-8 = dt
We change variables
y-8 = u
dy = du
We replace and integrate
∫ du / u = ∫ dt
Ln (y-8) = t
We evaluate at the lower limits t = 0 for y = y₀
ln (y-8) - ln (y₀-8) = t-0
Let's simplify the equation
ln (y-8 / y₀-8) = t
y-8 / y₀-8 =
y-8 = (y₀-8)
b) the equation is
dy / dt = 2y -5
u = 2y -5
du = 2 dy
du / 2u = dt
We integrate
½ Ln (2y-5) = t
We evaluate at the limits
½ [ln (2y-5) - ln (2y₀-5)] = t
Ln (2y-5 / 2y₀-5) = 2t
2y -5 = (2y₀-5)
c) the equation is very similar to the previous one
u = 2y -10
du = 2 dy
∫ du / 2u = dt
ln (2y-10) = 2t
We evaluate
ln (2y-10) –ln (2y₀-10) = 2t
2y-10 = (2y₀-10)
