In this question, it is given that the measurement of angles USW and TSR are 7x-34 and 4x+9 .
First we have to find the relationship between those two angles .
And these angles are vertical opposite angles which are congruent .
Therefore
Therefore measurement of angle USW is
Question 14
5 : 8
1.5 : x
Answer : Actual length = 2.4m
Question 15
5 : 8
x : 20
Answer : Actual length = 12.5m
Hope this helps. - M
Answer: 15.7oz < M < 16.3oz
Step-by-step explanation:
The average mass of each bag will be 16oz
If the bag weighs 0.3 oz more or less than 16 oz, then it is rejected.
Then the extremes are:
16oz - 0.3oz = 15.7oz (this mas counts as a rejected one)
So we have that the mass of the bag must be larger than 15.7oz
M > 15.7oz
Now, the other extreme is:
16oz +0.3oz = 16.3oz (again, this a rejected mass)
Then the mass of the bag must be smaller than 16.3oz
M < 16.3oz
Then the range of the accepted mass of the bags is:
15.7oz < M < 16.3oz
Answer:
Step-by-step explanation:
tan 3x=\frac{\sqrt{3} }{3} =\frac{1}{\sqrt{3} } =tan\frac{\pi }{6} =tan (n\pi +\frac{\pi }{6} )\\3x=n\pi +\frac{\pi }{6} =\frac{(6n+1)\pi }{6} \\x=\frac{(6n+1)\pi }{18} \\
where~x~is~an~integer.
A.) Let the length of the sides of the bottom of the box be y and z, and let the length of the sides of the square cut-outs be x, then
V = xyz . . . (1)
2x + y = 24 => y = 24 - 2x . . . (2)
2x + z = 24 => z = 24 - 2x . . . (3)
Putting (2) and (3) into (1), gives:
V = x(24 - 2x)(24 - 2x) = x(24 - 2x)^2 = x(576 - 96x + 4x^2)
V = 4x^3 - 96x^2 + 576x
b.) For maximum volume, dV/dx = 0
dV/dx = 12x^2 - 192x + 576 = 0
x^2 - 16x + 48 = 0
(x - 4)(x - 12) = 0
x = 4 or x = 12
but x = 12 is unrearistice
Therefore, x = 4.
y = z = 24 - 2(4) = 24 - 8 = 16
Therefore, the dimensions of the box that enclose the largest possible volume is 16 inches by 16 inches by 4 inches.
c.) Maximum volume = 16 x 16 x 4 = 1024 cubic inches.
