Answer:
negative
Step-by-step explanation:
The graph falls to both the left and right, so the leading coefficient has to be negative, and it must be raised to an even power.
Answer:2/16
Step-by-step explanation:
Answer:
Option C, 14/21=9/21 is the proportion.
Step-by-step explanation:
Given:
a.5/7 = 10/12
b.9/15 = 12/18
c.4/6 = 8/12
d.14/21 = 9/12
We have to find the proportions.
In proportion two ratios are equal, as it has an equality sign in between so both side must be of same ratio in its simplest form.
Let's work with option C
⇒ 
To find the simplest form we have to divide the numerator and denominator with same digit (or its factor).
Simplest form of
=
,
= 
Simplest form of
= 
Both sides in option C have equal ratios of 2/3.
So 4/6=8/12 is in proportion.
Question 11a)
We are given side BC equals to side CE and angle CBA equals to angle CED
We also know that angle ACB equals to angle ECD are equal (opposite angles properties)
We have enough information to deduce that triangle ABC and triangle CDE are equal by postulate Angle-Side-Angle (ASA)
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Question 11b)
We are given side AB equal to side ED, side BC equals to side EF, and side AC equals to side DF
We have enough information to deduce that triangle ABC and triangle DEF congruent by postulate Side-Side-Side (SSS)
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Question 11c)
We are given side AC equals to side DF, angle ABC equals to angle DEF, and angle BAC equals to angle EDF
We have enough information to deduce that triangle ABC congruent to triangle DEF by postulate Angle-Side-Angle (ASA)
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Question 11d)
We do not have enough information to tell whether this shape congruent or not
Remember: We have to work from either the LHS or the RHS.
(Left hand side or the Right hand side)
You should already know this:

You should also know this:

So plugging in both of those into our identity, we get:

Simplify the denominator on the LHS (Left Hand Side)
We get:

LHS = RHS
Therefore, identity is verified.