Find the circumference and round to the nearest hundredth
2 answers:
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The triangles that are similar would be ΔGCB and ΔPEB due to Angle, Angle, Angle similarity theorem.
<h3>How to identify similar triangles?</h3>From the image attached, we see that we are given the Parallelogram GRPC. Thus;
A. The triangles that are similar would be ΔGCB and ΔPEB due to Angle, Angle, Angle similarity theorem.
B. The proof of the fact that ΔGCB and ΔPEB are similar pairs of triangles is as follow;
∠CGB ≅ ∠PEB (Alternate Interior Angles)
∠BPE ≅ ∠BCG (Alternate Interior Angles)
∠GBC ≅ ∠EBP (Vertical Angles)
C. To find the distance from B to E and from P to E, we will first find PE and then BE by proportion;
225/325 = PE/375
PE = 260 ft
BE/425 = 225/325
BE = 294 ft
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Answer:
Step-by-step explanation:
The area of a triangle is
and we have the base (12m) and we need to find the height. Using Pythagorean's Theorem:
That's the height that we will sub into the area formula:
Answer:
the price of good x in 1999 dollars is 370.89 dollars
Step-by-step explanation:
Given that good x sold for $40 in 1945. the Cpi in 1945 was 18.0 and the cpi in 1999 was 166.6.
We have cpi and sale price have direct variation
In other words S = kC where C = CPi and S = sales price
In 1945, 40 = 18k or K = 20/9
Using this we can say
Sales price in 1999 would be k (166.6)
=
the price of good x in 1999 dollars is 370.89 dollars
Answer:
d = 234.6 m
Step-by-step explanation:
You can consider a system of coordinates with its origin at the beginning of the walk of the student.
When she start to walk, she is at (0,0)m. After her first walk, her coordinates are calculated by using the information about the incline and the distance that she traveled:
she is at the coordinates (52.97 , 28.16)m.
Next, when she walks 180m to the east, her coordinates are:
(52.97+180 , 28.16)m = (232.97 , 28.16)m
To calculate the distance from the final point of the student to the starting point you use the Pythagoras generalization for the distance between two points:
The displacement of the student on her complete trajectory was of 234.6m