Answer: If the wall is 300 meters long, the dimensions of the fence with the maximum area will be 300 × 150 = 45000 m²
Step-by-step explanation:
try combinations in a chart
190 × (600-190)/2 = 38950 = 190 × 205 increasing area
200 × (600-200)/2 = 40000 = 200 × 200
220 × (600-220)/2 = 41800 == 220 × 190
240 × (600-240)/2 = 43200 == 240 × 180
260 × (600-260)/2 = 44200 == 260 × 170
280 × (600-280)/2 = 44800 . == 280 × 160
<u>300 × (600-300)/2 = 45000 == 300 × 150</u> Maximum Area
310 × (600-310)/2 = 44950 . == 310 × 145 decreasing area
320 × (600-320)/2 = 44800 . == 320 × 140
The points define a trapezoid of height 6 and midsegment length 6. The area is
6×6 = 36 units²
Answer:
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Answer:
m∠2 = 78°
Step-by-step explanation:
From the given picture,
lines 'm' and 'n' are the parallel lines and line 't' is the transverse line.
∠1 and angle with the measure of 78° are corresponding angles.
Therefore, m∠1 = 78°
m∠1 = m∠2 [Vertical angles are equal]
m∠2 = 78°
Answer: nickels = 13, dimes = 7, quarters = 14
<u>Step-by-step explanation:</u>
First, set up the equations for each coin:

Next, the sum of the coins is $4.85 so substitute and solve for the variable:
Nickels + Dimes + Quarters = Sum
(0.30 + 0.05d) + 0.10d + 0.50d = 4.85
0.30 + 0.65d = 4.85
0.65d = 4.55
d = 7
Lastly, plug the d-value into the quantity equation for the nickels and quarters to find their quantity.
nickels: 6 + d = 6 + (7) = 13
quarters: 2d = 2(7) = 14