<span>−<span>1/<span><span>(<span>y+1</span>)</span><span>(<span>y<span>−1</span></span>) is the answer</span></span></span></span>
Answer:
We can not solve for a unique cost for each soldier.
Step-by-step explanation:
Let x be the daily cost for legionaries and y be the daily cost for archers.
Upon using our given information we will get a system of linear equations as:
![3x+3y=10...(1)](https://tex.z-dn.net/?f=3x%2B3y%3D10...%281%29)
![x+y=3...(2)](https://tex.z-dn.net/?f=x%2By%3D3...%282%29)
Now we will solve for x from our 2nd equation,
![x = 3-y](https://tex.z-dn.net/?f=x%20%3D%203-y)
Now we will substitute this value in our 1st equation.
![3(3-y)+3y=10](https://tex.z-dn.net/?f=3%283-y%29%2B3y%3D10)
We can see that -3y cancels out with 3y and 9 is not equal to 10. So this is an unsolvable system. Therefore, we can not find a unique cost for each soldier.
Answer:
y-30 = 3(x-1)
Step-by-step explanation:
Point-slope form of an equation is
y-y1 = m(x-x1) where (x1,y1 ) is a point and m is the slope
We know on day 1 they exercised 30 minutes (1,30) where x is the number of days and y is the minutes. The rate increased 3 minutes each day so m = 3
y-30 = 3(x-1)
Answer:
A = 20sinθ(6 + 5 cosθ) cm²
Step-by-step explanation:
Drop perpendiculars DE and CF to AB.
Then, we have congruent triangles ADE and BCF, plus the rectangle CDEF.
The formula for the area of the trapezium is
A = ½(a + b)h
DE = 10sinθ
AE = 10cosθ
BF = 10cosθ
EF = CD = 12 cm
AB = AE + EF + BF = 10cosθ + 12 + 10 cosθ = 12 + 20cosθ
A = ½(a + b)h
= ½(12 +12 + 20 cosθ) × 10 sinθ
=(24 + 20 cosθ) × 5 sinθ
= 4(6 + 5cosθ) × 5sinθ
= 20sinθ(6 + 5 cosθ) cm²
Answer:
Step-by-step explanation:
Multiplying x+6y=-3 with 2 so its equal to first eqn,
Eliminating,
2x + 3y = 3 (1)
2x +12y = -6 (2) (because subtracting)
------------------
-9y =9
∴y=-1
eq. y in 1
2x -3=3
2x=3+3
x=6/2
=3
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