Answer:
7. x=3, 8. x=7, 9. x=15.
Step-by-step explanation:
7. If lines m and n are congruent, then angles DCF and CFE are congruent. 15x+3=18x-6. Solve for x. --> 15x+9=18x-->3x=9-->x=3
8.If line m is parallel to line n, then the corresponding angles are congruent. So, 20x+1=22x-13 Solve for x. 20x+14=22x-->14=2x-->x=7.
9. The supplementary angle of 110 is 70. Note that all of the inner angles of a triangle are equal to 180. Form an equation using the the values: (4x+8)+(2x+12)+70=180. Simplify; 6x+90=180. Solve for x: 6x+90=180-->6x=90-->x=15.
Hope this helps!
Given:
second term = 18
fifth term = 144
The nth term of a geometric sequence is:

Hence, we have:

Divide the expression for the fifth term by the expression for the second term:

Substituting the value of r into any of the expression:

Hence, the explicit rule for the sequence is:
Answer:
The change to the face 3 affects the value of P(Odd Number)
Step-by-step explanation:
Analysing the question one statement at a time.
Before the face with 3 is loaded to be twice likely to come up.
The sample space is:

And the probability of each is:








P(Odd Number) is then calculated as:


Take LCM



After the face with 3 is loaded to be twice likely to come up.
The sample space becomes:

The probability of each is:








Take LCM


Comparing P(Odd Number) before and after
--- Before
--- After
<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>
Answer: A is the correct answer