The perimeter of the triangle is: (9.5n - 11.6p - 3.5q) cm.
<h3>What is the Perimeter of a Triangle?</h3>
The total length of all the sides of a triangle is equal to the perimeter of the triangle.
Given a triangle has the following lengths:
- (2.9n-7.8p) centimeters,
- (6.6n-6.4q) centimeters,
- (2.9q-3.8p) centimeters.
The perimeter of the triangle = (2.9n-7.8p) + (6.6n-6.4q) + (2.9q-3.8p)
The perimeter of the triangle = 2.9n - 7.8p + 6.6n - 6.4q + 2.9q - 3.8p
Combine like terms together
The perimeter of the triangle = 2.9n + 6.6n - 7.8p - 3.8p - 6.4q + 2.9q
The perimeter of the triangle = 9.5n - 11.6p - 3.5q
Thus, the perimeter of the triangle is: (9.5n - 11.6p - 3.5q) cm.
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Josiah and Chana travel at constant and different speeds.
- The point F indicates that after 25 seconds Josiah is 60 meters from the starting line but behind Chana
<h3>How can the what the the point <em>F </em>represent be known?</h3>
Josiah's head start = 10 meters from the start
Josiah's speed = 2 m/s
Chana's speed = 3 m/s
Expressing the distance traveled as an equation, we have;
D = d + s × t
Where;
D = The distance covered
d = The distance from the starting line the runner starts
s = The speed of the runner
t = The time spent running
For Josiah, we have;
D = 10 + 2•t (line <em>a</em>)
For Chana, we have;
D = 0 + 3•t = 3•t (line <em>b</em>)
The above equations are straight line equations.
The point <em>F </em>is on line <em>a</em>, which shows Josiah distance after 25 seconds which is 60 meters. The corresponding point on line <em>b</em>, Chana's distance after 25 minutes is 75 meters.
Therefore;
- The point <em>F </em>indicates that after 25 seconds Josiah is 60 meters from the starting line but behind Chana
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Do you think the answer is C.147?
For T = 89.5 degrees C:
E(t) = 1100(90 - 89.5) + 520(90 - 89.5)^2= 1100(0.5) + 520(0.5)^2= 550 + 130= 680 m
For T = 90 degrees C:
E(t) = 1100(90 - 90) + 520(90 - 90)^2= 0 + 0= 0 m (meaning that boiling point is 90 degrees C at ground level)
Answer:
All real numbers is the answer
Step-by-step explanation: