Answer:
-2/3
Step-by-step explanation:
2x + 3y = - 3 re-arrange to y = mx+b form: slope of this line will be 'm'
y = -2/3 x -1 slope = -2/3 parallel slope is the same
Answer:
Step-by-step explanation:
your equation shown above is set up incorrectly in order to solve this i need the right equation you must plug in each number to q to find the right solution
Good evening ,
______
Answer:
1) (f+g)(1) = e
2) (fg)(1) = 0
3) (3f)(1) = 3e
___________________
Step-by-step explanation:
1) (f+g)(1) = f(1) + g(1) = e¹ + log1 = e + 0 = e.
2) (fg)(1) = f(1) × g(1) = e¹ × log(1) = e¹ × 0 = e × 0 = 0.
3) (3f)(1) = 3×f(1) = 3×e¹ = 3e.
:)
Let x=ab=ac, and y=bc, and z=ad.
Since the perimeter of the triangle abc is 36, you have:
Perimeter of abc = 36
ab + ac + bc = 36
x + x + y = 36
(eq. 1) 2x + y = 36
The triangle is isosceles (it has two sides with equal length: ab and ac). The line perpendicular to the third side (bc) from the opposite vertex (a), divides that third side into two equal halves: the point d is the middle point of bc. This is a property of isosceles triangles, which is easily shown by similarity.
Hence, we have that bd = dc = bc/2 = y/2 (remember we called bc = y).
The perimeter of the triangle abd is 30:
Permiter of abd = 30
ab + bd + ad = 30
x + y/2 + z =30
(eq. 2) 2x + y + 2z = 60
So, we have two equations on x, y and z:
(eq.1) 2x + y = 36
(eq.2) 2x + y + 2z = 60
Substitute 2x + y by 36 from (eq.1) in (eq.2):
(eq.2') 36 + 2z = 60
And solve for z:
36 + 2z = 60 => 2z = 60 - 36 => 2z = 24 => z = 12
The measure of ad is 12.
If you prefer a less algebraic reasoning:
- The perimeter of abd is half the perimeter of abc plus the length of ad (since you have "cut" the triangle abc in two halves to obtain the triangle abd).
- Then, ad is the difference between the perimeter of abd and half the perimeter of abc:
ad = 30 - (36/2) = 30 - 18 = 12
The answer is no solution.
First you simplify both sides of the equation... Then you subtract 6x from both sides. And lastly you subtract 6 from both sides. Leaving you with no solutions.
