This is false. An arbitrary number is the same as a random number. It is any number that is not specified, not restricted and not consciously thought of. Simply put, an arbitrary number or a random number is any number that you can think of.
Answer:
2 pi, -2 pi, pi, -pi
Step-by-step explanation:
Draw a unit circle and label the 4 points where it intersects the x/y axis. for example at 0 degrees or 360/-360 degrees the (x,y) or (cosine theta, sine theta) value is (1,0).
Perimeter = 2(width+length) =28
Let x=width of base, then length of base = (28/2)-x = 14-x
Volume of box = length*width*height
=x(14-x)*5 = 5x(14-x)
The answer is 12>x>-7
Step-by-step explanation:
it's first divided into 2 parts
1 . 15 > 2x - 9
2. 2x - 9 > -23
Then solve them individually.
let's do it,
<u>First half</u>
15 > 2x - 9
15+9 > 2x
24 > 2x
24/2 > x
12>x
<u>Second half</u>
2x-9>-23
2x>-23+9
2x>-14
x>-14/2
x>-7
<u>Joining of the two halves</u>
We have :
12>x>-7
X=2h, y=3k
Substitute these values into equations.
y+2x = 4 ------> 3k+2*2h=4 -----> 3k +4h =4
2/y - 3/2x = 1-----> 2/3k -3/(2*2h) = 1 ------> 2/3k - 3/4h =1
We have a system of equations now.
3k +4h =4 ------> 3k = 4-4h ( Substitute 3k in the 2nd equation.)
2/3k - 3/4h =1
2/(4-4h) -3/4h = 1
2/(2(2-2h)) - 3/4h = 1
1/(2-2h) -3/4h - 1=0
4h/4h(2-2h) -3(2-2h)/4h(2-2h) - 4h(2-2h)/4h(2-2h) =0
(4h- 3(2-2h) - 4h(2-2h))/4h(2-2h) = 0
Numerator should be = 0
4h- 3(2-2h) - 4h(2-2h)=0
Denominator cannot be = 0
4h(2-2h)≠0
Solve equation for numerator=0
4h- 3(2-2h) - 4h(2-2h)=0
4h - 6+6h-8h+8h² =0
8h² +2h -6=0
4h² + h-3 =0
(4h-3)(h+1)=0
4h-3=0, h+1=0
h=3/4 or h=-1
Check which
4h(2-2h)≠0
1) h= 3/4 , 4*3/4(2-2*3/4)=3*(2-6)= -12 ≠0, so we can use h= 3/4
2)h=-1, 4(-1)(2-2*(-1)) =-4*4=-16 ≠0, so we can use h= -1, also.
h=3/4, then 3k = 4-4*3/4 =4 - 3=1 , 3k =1, k=1/3
h=-1, then 3k = 4-4*(-1) =8 , 3k=8, k=8/3
So,
if h=3/4, then k=1/3,
and if h=-1, then k=8/3 .
