Answer:
x=-2.25,y=-2.25
Step-by-step explanation:
The given system is
We want to solve the system for x and y that makes the system true.
Let us multiply the first equation by 3 while maintaining the second equation:
Subtract the current second equation from the first one.
Divide through by 12:
From the second equation we have:
This implies that:
Answer:
A .cos(x)<1
Step-by-step explanation:
According to the first inequality
cos(x)<1
x < arccos 1
x<0
This therefore does not have a solution within the range 0 ≤ x ≤ 2pi
x cannot be leas than 0. According to the range not value, 0≤x which is equivalent to x≥0. Thus means otvis either x = 0 or x> 0.
For the second option
.cos(x/2)<1
x/2< arccos1
x/2<0
x<0
This inequality also has solution within the range 0 ≤ x ≤ 2pi since 0 falls within the range of values.
For the inequality csc(x)<1
1/sin(x) < 1
1< sin(x)
sinx>1
x>arcsin1
x>90°
x>π/2
This inequality also has solution within the range 0 ≤ x ≤ 2pi since π/2 falls within the range of values
For the inequality csc(x/2)<1
1/sin(x/2) < 1
1< sin(x/2)
sin(x/2)> 1
x/2 > arcsin1
X/2 > 90°
x>180°
x>π
This value of x also has a solution within the range.
Therefore option A is the only inequality that does not have a solution with the range.
In rectangle ABCD
AB * AC = area
In another rectangle WXYZ
WX * WY = area
The sides must be equal or have same multiples and should be divisible by each other
L ≥ 5w + 4
2(L+w) ≥ 32
L+w ≥ 16
6w + 4 ≥ 16
6w ≥ 12
w ≥ 2
L ≥ 5w + 4
L ≥ 14
If L = 14 and w = 2 then the perimeter is 32.
These are the smallest dimensions that fit the criteria.
Note: If we raise the width by 1 to w = 3
then the length is at least 19. At 19 by 3 we have P = 44
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