Answer: x=8
Step-by-step explanation:
the sum of the interior angles of a quadrilateral is 360 (number of sides-2 * 180)
so

<h3>
Answer: 80 degrees</h3>
============================================================
Explanation:
I'm assuming that segments AD and CD are tangents to the circle.
We'll need to add a point E at the center of the circle. Inscribed angle ABC subtends the minor arc AC, and this minor arc has the central angle AEC.
By the inscribed angle theorem, inscribed angle ABC = 50 doubles to 2*50 = 100 which is the measure of arc AC and also central angle AEC.
----------------------------
Focus on quadrilateral DAEC. In other words, ignore point B and any segments connected to this point.
Since AD and CD are tangents, this makes the radii EA and EC to be perpendicular to the tangent segments. So angles A and C are 90 degrees each for quadrilateral DAEC.
We just found angle AEC = 100 at the conclusion of the last section. So this is angle E of quadrilateral DAEC.
---------------------------
Here's what we have so far for quadrilateral DAEC
- angle A = 90
- angle E = 100
- angle C = 90
- angle D = unknown
Now we'll use the idea that all four angles of any quadrilateral always add to 360 degrees
A+E+C+D = 360
90+100+90+D = 360
D+280 = 360
D = 360-280
D = 80
Or a shortcut you can take is to realize that angles E and D are supplementary
E+D = 180
100+D = 180
D = 180-100
D = 80
This only works if AD and CD are tangents.
Side note: you can use the hypotenuse leg (HL) theorem to prove that triangle EAD is congruent to triangle ECD; consequently it means that AD = CD.
Answer:
20, 4
Step-by-step explanation:
Let's assume total number of students registered is n
Last week n/4 people are absent
so people present are n - n/4 = <u>3n/4</u>
Today after return of Jen 1/5 n are absent
so people present are n - n/5 = 4n/5
which is also equal to last week present people + 1 (Jen)
= 3n/4 + 1
= (3n + 4) / 4
so (3n + 4) / 4 = 4n/5
=> 3n + 4 = 4n/5 * 4
=> (3n + 4) * 5 = 16n
=> 15n + 20 = 16n
=> 20 = n
=> n = 20
So total students registered for class is 20
today no of people absent = n/5 = 20/5 = 4
Answer:
he can run 5 meters in 1 min
Step-by-step explanation:
S = d/t
S=100m/20s
S=5m/s
I'm not sure I understand the question. Is this worded correctly?