The correct answer is p/rs.
In order to find this first state the direct variation with the constant k included.
p = krs
Now to find the value of k, we simply divide away the r and s
p/rs = k
Answer:
90.80
Step-by-step explanation:
6x+4
= (6)(13)+4
= 78+4
=82
3x
=(3)(13)
= 39
So we know the sides are 39 and 82
Pythagoras theorem in triangles
= a2+b2= c2
Now, we know the value of a and B but not c (the hypotenuse)
Therefore,
c2 = (39)^2 + (82)^2
= 1521+ 6724
= 8245
so, c = √8245
= 90.80 unitd
The answer is B because on the graph you start at $3.00 then you add $5.00 per hour so it goes up 5/1 per hour on the graph.
Answer:
b. both functions have a negative slope
Step-by-step explanation:
The given function is

This function is of the form
, where
is the slope and
is the y-intercept.
We can also determine the slope of the function in the table using any two points, say
.
The slope formula is given by,
.
If
, then,

.
We can see that both functions have a negative slope.
The correct answer is B.
Answer:
For tingle #1
We can find angle C using the triangle sum theorem: the three interior angles of any triangle add up to 180 degrees. Since we know the measures of angles A and B, we can find C.



We cannot find any of the sides. Since there is noting to show us size, there is simply just not enough information; we need at least one side to use the rule of sines and find the other ones. Also, since there is nothing showing us size, each side can have more than one value.
For triangle #2
In this one, we can find everything and there is one one value for each.
- We can find side c
Since we have a right triangle, we can find side c using the Pythagorean theorem






- We can find angle C using the cosine trig identity




- Now we can find angle A using the triangle sum theorem



For triangle #3
Again, we can find everything and there is one one value for each.
- We can find angle A using the triangle sum theorem



- We can find side a using the tangent trig identity




- Now we can find side b using the Pythagorean theorem



