17.5% leaving load
You do 0.0175 divided by the money
$2937.50
———-
0.0175
This kind of exercise can be real drudgery. But it's almost all simple arithmetic, so better you than me. I'll do one of these for you, which will show you how to do the other one.
a). sin · cot / sec
You're supposed to know that [ cotangent = cosine/sine ]
and [ secant = 1/cosine ].
Then the problem becomes
sin · (cos/sin) / (1/cos) = cos²
Aw shucks, I might as well also set up 'b)' for you:
b). cos · csc / tan
You're supposed to know that [ cosecant = 1/sine ]
and [ tangent = sine/cosine ].
Then the problem becomes
cos · (1/sin) / (sin/cos) = (cos/sin)²
Now simply plug in the given values of cos A and sin A .
And may I compliment you on your nail care !
The longest possible altitude of the third altitude (if it is a positive integer) is 83.
According to statement
Let h is the length of third altitude
Let a, b, and c be the sides corresponding to the altitudes of length 12, 14, and h.
From Area of triangle
A = 1/2*B*H
Substitute the values in it
A = 1/2*a*12
a = 2A / 12 -(1)
Then
A = 1/2*b*14
b = 2A / 14 -(2)
Then
A = 1/2*c*h
c = 2A / h -(3)
Now, we will use the triangle inequalities:
2A/12 < 2A/14 + 2A/h
Solve it and get
h<84
2A/14 < 2A/12 + 2A/h
Solve it and get
h > -84
2A/h < 2A/12 + 2A/14
Solve it and get
h > 6.46
From all the three inequalities we get:
6.46<h<84
So, the longest possible altitude of the third altitude (if it is a positive integer) is 83.
Learn more about TRIANGLE here brainly.com/question/2217700
#SPJ4
(3,2)(0,0)
slope = (2-0)/(3-0) = 2/3
b = 0
so
equation
y = 2/3x
answer is C.
y = 2/3x
