Answer:
P=40
a + b + x = P
a = 3x
b = x + 5
substitute for b & y
3x + (x+5) + x = 40
Step-by-step explanation:
longest section 1 = a
mid length section 2 = b
shortest section 3 = x
P=40
a + b + x = P
a = 3x
b = x + 5
substitute for b & y
3x + (x+5) + x = 40
5x + 5 = 40
5x = 35
x = 7
b = 12
a = 21
The number of bars which can be cut for the 1" X 1" and 2" X 2" square bars are; 117 and 29 units respectively.
<h3>How many bars can be cut from the pan in each case?</h3>
The total area of the given pan can be evaluated as follows;
Area = length × width
= 9 × 13
= 117 square units.
Hence, the number of 1" X 1" bars can be cut which can be cut from the pan;
= 117/(1×1)
= 117 1" X 1" bars.
For the 2" X 2" square bars, we have;
= 117/(2×2);
29 remainder 1 square unit.
Ultimately, one square unit of the pan is wasted for the 2" X 2" square bars.
Read more on area;
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Answer:
Step-by-step explanation:
130 decrease
Option B: 1.6 s.
Although the equation is not included, you can obtain it by physical considerations
H(t) = -16.t^2 + 52t + 16
The maximum point is at the vertex and the vertex is in the middle of the roots.
The roots are when H(t) = 0,
So you have to solve 0 =-16t^2 + 52t + 16
Use the quadratic formula and you will find t = 3.53 and t =-0.28
Then the vertex is at the middle point = (3.53 -.028)/2 =1.63 s and the answer is option B: 1.6 s.
