Answer:
i did the math in my calculator and came up with -4550 i don't know what i did
Step-by-step explanation:
Answer:
Her adjusted body weight is 78.26kg = 78.3kg, rounded to the nearest tenth.
Step-by-step explanation:
To find the Adjusted Body Weight(AjBW), first we have to consider the Ideal Body Weight(IBW).
We have the following formulas:
, in which i is the number of inches that the woman has above 5 feet.
, in which
is her Actual Body Weight
So, in this problem:
Each pound has 0.45kg. So her weight is
kg.
The woman is 8 inches above 5 feet, so
.

kg
Her ideal body weight is 63.9kg. Her adjusted body weight is:


kg
Her adjusted body weight is 78.26kg = 78.3kg.
Tanα=h/x
x=h/tanα
x=70/tan38 ft
x≈89.6 ft (to nearest tenth of a foot)
Note: the question seems strange, (because this would make the ladder almost 114 feet long!) or are we to assume as I did that the ladder is leaning against a point 70 feet high, or should if be a 70 foot ladder leaning against a wall of unknown height?
If it is just a 70 ft ladder at a 38° inclination the:
cosα=x/L
x=Lcosα
x=70cos38° ft
x≈55.2 ft
Area of the 3 rectangles that make up the prism:
30 * 25 + 30 * 17 + 30 * 28 = 2100
Area of the two triangles at either end:
Use bh/2
(15 * 28)/2 = 210
There are two so it’s 420
2100 + 420 = 2520cm^2