The easiest way to attempt such questions is to throw in some numbers and do trial and error.
The question is asking about the change in values
depending on the change in
. So, let's play around with
.
Take
, then
Take
, then
Take
, then
.
.
.
Take
, then
Notice how
decreases in value for every increase in
?
Therefore, as
increases, the value of
decreases (always).
Ans: B
Answer:
He gave 20
Step-by-step explanation:
30 minus 10 equals 20 and if John gave times that he would have 20 (40-20) which will give him the same answer.
Problem 1
<h3>Answers:</h3><h3>arc AB = 122 degrees</h3><h3>angle ADB = 61 degrees</h3>
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Explanation:
Arc BC is 58 degrees. The arc AB combines with it to form semicircle ABC, which is basically arc ABC with measure 180
This must mean arc AB is 180-58 = 122 degrees
When in doubt, chances are the number 180 might be involved (in some way) when it comes to geometry problems. Of course, this isn't always a guarantee. The number does show up a lot though.
Once we determine what arc AB is equal to, we cut it in half to get the inscribed angle ADB. Refer to the inscribed angle theorem. Notice how angle ADB subtends the arc AB.
==========================================================
Problem 2
<h3>Answer: 9</h3>
--------------------
Explanation:
x = length of AE
Since AC = 13, this makes EC = 13-x
We'll use the intersecting chords theorem to find x.
(AE)*(EC) = (BE)*(DE)
x*(13-x) = 6*6
13x-x^2 = 36
-x^2 + 13x - 36 = 0
x^2 - 13x + 36 = 0
(x-9)(x-4) = 0
x-9 = 0 or x-4 = 0
x = 9 or x = 4
This means AE could be 9 units long or 4 units long.
Usually the diagram isn't drawn to scale so we don't really use it all too much (other than initial set up). However, if the diagram was drawn to scale, then notice that AE is over the halfway point of AC. This means we would be able to rule out x = 4 since 4 is not over the halfway point of 13. I have a feeling this is what your teacher might be going for.
In short, only AE = 9 is possible.
X = inside angle
x = 86
x = 86 degrees
9.00/315 is a way to solve this.
