Answer:
No
Step-by-step explanation:
You have to build the triangles.
They are such that:
h is the common height
x is the horizontal distance from the plane to one stone
Beta is the angle between x and the hypotenuse
Then in this triangle: tan(beta) = h / x ......(1)
1 - x is the horizontal distance from the plane to the other stone
alfa is the angle between 1 - x and h
Then, in this triangle: tan (alfa) = h / [1 -x ] ...... (2)
from (1) , x = h / tan(beta)
Substitute this value in (2)
tan(alfa) = h / { [ 1 - h / tan(beta)] } =>
{ [ 1 - h / tan(beta) ] } tan(alfa) = h
[tan(beta) - h] tan(alfa) = h*tan(beta)
tan(beta)tan(alfa) - htan(alfa) = htan(beta)
h [tan(alfa) + tan(beta) ] = tan(beta) tan (alfa)
h = tan(beta)*tan(alfa) / (t an(alfa) + tan(beta) )
B = 100(1 + 0.04)^12 = 100(1.04)^12 = 100(1.601) = $160.10
Answer:
2 units up for every one unit across, or just simply 2
Step-by-step explanation:
Here we must see in how many different ways we can select 2 students from the 3 clubs, such that the students <em>do not belong to the same club. </em>We will see that there are 110 different ways in which 2 students from different clubs can be selected.
So there are 3 clubs:
- Club A, with 10 students.
- Club B, with 4 students.
- Club C, with 5 students.
The possible combinations of 2 students from different clubs are
- Club A with club B
- Club A with club C
- Club B with club C.
The number of combinations for each of these is given by the product between the number of students in the club, so we get:
- Club A with club B: 10*4 = 40
- Club A with club C: 10*5 = 50
- Club B with club C. 4*5 = 20
For a total of 40 + 50 + 20 = 110 different combinations.
This means that there are 110 different ways in which 2 students from different clubs can be selected.
If you want to learn more about combination and selections, you can read:
brainly.com/question/251701
