Answer:
94 for the first one and 130 for the second
Step-by-step explanation:
Since A=6 and B=4
For the first one
13(6)+ 4(4)
78+16=94
For the second one
(6+4)*13
(10)*13= 130
The algebraic expression n/6 represents that the class was divided into 6 equal groups.
Answer:
15. 12
Step-by-step explanation:
just multiply them
Because the cosine and sine must be negative when evaluated in theta, the angle lies on the third quadrant.
<h3>
In which quadrant is the endpoint of the segment that defines the angle?</h3>
We know that if:
- cos(θ) > 0, then we are on the first or fourth quadrant
- sin(θ) > 0, then we are on first or second quadrant.
Here we know that:
sec(θ) < 0
And we know that:
sec(θ) = 1/cos(θ)
Then we have cos(θ) < 0
We also have that:
sec(θ)*csc(θ) > 0
Because sec(θ) < 0, we must have that csc(θ) < 0.
Remember: csc(θ) = 1/sen(θ)
Then sen(θ) < 0.
Then we have the two conditions:
sen(θ) < 0
cos(θ) < 0
- The cosine is negative on the third and second quadrants.
- The sine is negative on the third and fourth quadrants.
The only quadrant where both are negative is the third quadrant, so that is the correct option.
If you want to learn more about trigonometric functions:
brainly.com/question/8120556
#SPJ1
M=log I / S
if M=6, 6=logI/S and then 10^6=I/S, so we get I=S <span>x 10^6, we assume that
</span>A=S x 10^6
if M=8, so, 8=logI/S and then 10^8=I/S, so we get I= S x 10^8=A=S x 10^6x10^2
since A=S x 10^6, so I= <span>A x10^2, </span><span>an earthquake with a magnitude of 8 is 10² stronger than an earthquake with a magnitude of 6.</span>
