X = width
2x + x = 3 · 13.73
3x = 41.19
x = 13.73
The width is 13.73 units.
They are similar. Both isosceles triangles
Answer:
option 4
Step-by-step explanation:
using the equation stated above
1st term = -6+(1-1=0)*(6)=-6
4th term= -6+(4-1=3)*6=12
10th term= -6+(10-1=9)*6=48
a ) The domain:
x^4 - 16 x² ≥ 0
x² ( x² - 16 ) ≥ 0
x² - 16 ≥ 0
x² ≥ 16
x ∈ ( - ∞, - 4 ] ∪ [ 4 , + ∞ )b ) f ` ( x ) =
=
( 2 x³ - 16 x ) / √(x^4 - 16 x²)c ) The slope of the tangent line at x = 5:
f ` ( 5 ) = ( 2 * 125 - 16 * 5 ) / √ ( 625 - 400 ) = 170 / 15 = 34 / 3
The slope of the line normal to the graph at x = 5:
m = - 3 / 34
Answer:
Step-by-step explanation:
Let's call hens h and ducks d. The first algebraic equation says that 6 hens (6h) plus (+) 1 duck (1d) cost (=) 40.
The second algebraic equations says that 4 hens (4h) plus (+) 3 ducks (3d) cost (=) 36.
The system is
6h + 1d = 40
4h + 3d = 36
The best way to go about this is to solve it by substitution since we have a 1d in the first equation. We will solve that equation for d since that makes the most sense algebraically. Doing that,
1d = 40 - 6h.
Now that we know what d equals, we can sub it into the second equation where we see a d. In order,
4h + 3d = 36 becomes
4h + 3(40 - 6h) = 36 and then simplify. By substituting into the second equation we eliminated one of the variables. You can only have 1 unknown in a single equation, and now we do!
4h + 120 - 18h = 36 and
-14h = -84 so
h = 6.
That means that each hen costs $6. Since the cost of a duck is found in the bold print equation above, we will sub in a 6 for h to solve for d:
1d = 40 - 6(6) and
d = 40 - 36 so
d = 4.
That means that each duck costs $4.
