Answer:
n = 95/14, or 6 11/14
Step-by-step explanation:
Let the number be n. Then:
n - 1 2/7 = 5 1/2.
Multiplying each term by the LCD (which is 14), we get:
14n - 18 = 77
Then 14n = 95
and so the desired result is n = 95/14, or 6 11/14
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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.
To ease your problem, consider "L" as you x-axis
Then the coordinate become:
A(- 4 , 3) and B(1 , 2) [you notice that just the y's changed]
This is a reflection problem.
Reflect point B across the river line "L" to get B', symmetric of B about L.
The coordinates of B'(1 , -1) [remember L is our new x-axis]
JOIN A to B' . AB' intersect L, say in H
We have to find the shortest way such that AH + HB = shortest.
But HB = HB' (symmetry about L) , then I can write instead of
AH + HB →→ AH + HB'. This is the shortest since the shortest distance between 2 points is the straight line and H is the point requiered
Free points rights anyways can someone help me with this answer
A 85-kg is clown is fired from rest out of 480-kg cannon. If the clowns speed is 18 meters per second after the firing, then what is the recoil speed of the cannon? (Round your answer to the nearest tenth)
