Answer:
Here we have two triangle rectangles, and we want to know the value of y, the shared cathetus between them.
First, we know that one of them has an angle of 60° and the adjacent cathetus to it is a.
The other has an angle of 30° and the adjacent cathetus is b.
In both cases, the opposite cathetus is y.
Now we can recall the relationship:
Tan(a) = (opposite cathetus)/(adjacent cathetus)
Then we can write:
tan(60°) = y/a
tan(30°) = y/b
And we also know that:
a + b = 15
Then we have 3 equations:
tan(60°) = y/a
tan(30°) = y/b
a + b = 15
To solve this, we first need to isolate one variable in one of the equations, i will isolate a in the third equation:
a = 15 - b
Now we can replace it in the first equation to get:
tan(60°) = y/(15 - b)
i will rewrite this one as:
tan(60°)*(15 - b) = y
then our system is:
tan(60°)*(15 - b) = y
tan(30°) = y/b
now we can isolate b in the second equation to get:
b = y/tan(30°)
and then replace it on the other equation to get:
tan(60°)*(15 - y/tan(30°)) = y
Now we can remember that:
tan(60°) = √3
tan(30°) = (√3)/3
Replacing these in our equation we get:
√3*(15 - y/((√3)/3)) = y
we can rewrite this as:
√3*(15 - y*3/√3) = y
then:
(√3)*15 - y*3 = y
(√3)*15 = y + y*3
(√3)*15 = 4*y
(√3)*15/4 = y
then we get:
X = Heather's age NOW
x - 2 = her age 2 years ago
x - 3 = her age 3 years ago
The Least Common Multiple of 2 and 3 is 6. Multiply both sides by 6.
Answer:
y= = -2
Step-by-step explanation:
In analytic geometry, an asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tends to infinity.
The answer is 112 Explanation: In order to divide
1
3
4
,
1st: you must find the reciprocal of the
denominator, which is
1
4
2nd: Then you multiply
1
3
by the reciprocal
So
1
3
4
will now be
1
3
x
1
4
=
1
12
Answer:
each van can transport 15 students
Step-by-step explanation:
186 divied by 2 and you get 93 then you divied that by 6 and you get 15.5 you cant have .5 sutdents on a van so you round down and get 15 students.