Answer:
Answer:
The graph will be
2 units away from the origin on positive
and three units upward from the origin towards
.
Step-by-step explanation:
Here is a graph attached with it.
To graph
we know that positive
is a
shaped from the origin.
Key points:
- To move rightward there must be a negative inside the parentheses.
- And to move upward we must have positive
.
If we have to move towards
then we must have negative inside it.
And if we have to move upward in
positive we must have positive constant value.
So the graph will be three units upward and two units rightward with a V-shaped ray.
Answer:
![7\sqrt{3}](https://tex.z-dn.net/?f=7%5Csqrt%7B3%7D)
Step-by-step explanation:
Since this is a 30-60-90 triangle, we know that the sides have the following characteristic:
The side opposite to 30 degree angle: n
The side opposite to 60 degree angle: ![n\sqrt{3}](https://tex.z-dn.net/?f=n%5Csqrt%7B3%7D)
The side opposite to 90 degree angle: 2n
Since we know that 7 is opposite to 30-degree, and x is opposite to 60 degree, than we know that x = ![7\sqrt{3}](https://tex.z-dn.net/?f=7%5Csqrt%7B3%7D)
Answer: A & C
<u>Step-by-step explanation:</u>
HL is Hypotenuse-Leg
A) the hypotenuse from ΔABC ≡ the hypotenuse from ΔFGH
a leg from ΔABC ≡ a leg from ΔFGH
Therefore HL Congruency Theorem can be used to prove ΔABC ≡ ΔFGH
B) a leg from ΔABC ≡ a leg from ΔFGH
the other leg from ΔABC ≡ the other leg from ΔFGH
Therefore LL (not HL) Congruency Theorem can be used.
C) the hypotenuse from ΔABC ≡ the hypotenuse from ΔFGH
at least one leg from ΔABC ≡ at least one leg from ΔFGH
Therefore HL Congruency Theorem can be used to prove ΔABC ≡ ΔFGH
D) an angle from ΔABC ≡ an angle from ΔFGH
the other angle from ΔABC ≡ the other angle from ΔFGH
AA cannot be used for congruence.
Answer:
2.2
Step-by-step explanation: