Let h units denote the hypotenuse of the smaller triangle. From the Pythagorean Theorem, we derive specific relationships involving the smaller triangle with dimensions
along with the shorter leg of the second triangle denoted as s units. Furthermore, we apply the double angle property and substitute values to arrive at the final calculation.
The options you provided are unclear to me, so I will respond in general terms: to determine if a point (ordered pair) lies on a line, you need to substitute the x-value from that pair into the line's equation and check if the resulting y-value matches the y-value of the ordered pair. For example, if your line is y = 4/3x + 1/3, we can check if (0, 0) and (2, 3) fit this line. We find that y = 4/3·0 + 1/3 gives us 1/3, which does not equal 0, indicating (0, 0) is not on the line. For (2, 3), substituting yields y = 4/3·2 + 1/3 = 3, meaning (2, 3) is on the line.
Answer:
The accurate assertions include:
A. m∠6 = 55°
C. m∠1 + m∠4 = 250°
D. m∠1 + m∠6 = m∠7 + m∠4
Step-by-step clarification:
The provided information is as follows:
m∠7 = 55°
The angles formed by the transversal and the upper horizontal parallel line are (starting from the top left and moving in clockwise direction) 1, 2, 4, 3
Likewise, the angles from the transversal and the lower horizontal parallel line are (starting from the top left and going clockwise) 5, 6, 8, 7
Consequently, we have;
m∠7 ≅ m∠6 (Vertically opposite angles are equal)
Thus, m∠6 = m∠7 = 55°
m∠6 = 55°, which aligns with option A.
m∠5 + m∠6 = 180° (The sum of angles on a straight line)
So, m∠5 = 180° - m∠6 = 180° - 55° = 125°
m∠5 = 125°
m∠1 ≅ m∠5 (Corresponding angles)
Thus, m∠1 = m∠5 = 125°
m∠1 ≅ m∠4 (Vertically opposite angles)
Therefore, m∠1 = m∠4 = 125°
Thus, m∠1 + m∠4 = 125° + 125° = 250°
m∠1 + m∠4 = 250°, which corresponds to option C.
m∠1 ≅ m∠4 (Vertically opposite angles)
m∠6 ≅ m∠7 (Vertically opposite angles)
Thus, m∠1 + m∠6 = m∠7 + m∠4 (Transitive property), which matches option D.
The salt enters at a rate of (5 g/L)*(3 L/min) = 15 g/min.
The salt exits at a rate of (x/10 g/L)*(3 L/min) = 3x/10 g/min.
Thus, the total rate of salt flow, represented by 
in grams, is defined by the differential equation,
which is linear. Shift the 
term to the right side, then multiply both sides by 
:
Next, integrate both sides and solve for :
Initially, the tank contains 5 g of salt at time 
, so we have
The duration required for the tank to contain 20 g of salt is 
, such that
1,107 cc
The scanning consists of 10 intervals:
[0,1.5), [1.5,3), [3,4.5), [4.5,6), [6,7.5), [7.5,9), [9,10.5), [10.5,12), [12,13.5), [13.5,15)
To estimate the volume using the Midpoint Rule, n should be set to 10.
Given that we will use n=5, we will split the range [0,15] into five intervals of lengths 3 each:
[0,3], [3,6], [6,9], [9,12], [12,15] and calculate their midpoints:
1.5, 4.5, 7.5, 10.5, and 13.5.
Next, we will determine the volume V from the five cylinders, where each has a height h=3 and the base area A corresponds to the calculated midpoints' intervals:
Cylinder 1
Midpoint=1.5, corresponding to the 2nd interval
A = 18, V= height * area of the base = 18*3 = 54 cc
Cylinder 2
Midpoint=4.5, corresponding to the 4th interval
A = 78, V= height * area of the base = 78*3 = 234 cc
Cylinder 3
Midpoint=7.5, corresponding to the 6th interval
A = 106, V= height * area of the base = 106*3 = 318 cc
Cylinder 4
Midpoint=10.5, corresponding to the 8th interval
A = 129, V= height * area of the base = 129*3 = 387 cc
Cylinder 5
Midpoint=13.5, corresponding to the 10th interval
A = 38, V= height * area of the base = 38*3 = 114 cc
Thus, the estimated volume is
54 + 234 + 318 + 387 + 114 = 1,107
