1. One face is 5.1*2.5=12.75 un^2, one is 8.5*2.5=21.25, one is 6.8*2.5=17, and two are 1/2*6.8*5.1, so both are 34.68. Total is 85.68, rounded to 85.7 cm^2. (C)
2. The area of the circles is 2(π*2^2), which is about 25.1. The rectangle's area can be found as 2π*2*3, which is about 37.7. The total is rounded to 62.8 m^2 (D)
3. Rectangular faces: 3.6*1.8=6.48, 8.5*1.8=15.3, 7.7*1.8=13.86
Triangles: 2(1/2(3.6*7.7))=27.72
Total: 63.36, round to 63.4 km^2 (B)
Answer:
53.3324
Step-by-step explanation:
given that a thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 40° F.
By Newton law of cooling we have
T(t) = ![T+(T_0-T)e^{-kt}](https://tex.z-dn.net/?f=T%2B%28T_0-T%29e%5E%7B-kt%7D)
where T (t) is temperature at time t,T =surrounding temperature = 40, T0 =70 = initial temperature
After half minute thermometer reads 60° F. Using this we can find k
![T(0,5) = 40+(70-40)e^{-k/2} = 60\\e^{-k/2} =2/3\\-k/2 = -0.4055\\k = 0.8110](https://tex.z-dn.net/?f=T%280%2C5%29%20%3D%2040%2B%2870-40%29e%5E%7B-k%2F2%7D%20%3D%2060%5C%5Ce%5E%7B-k%2F2%7D%20%3D2%2F3%5C%5C-k%2F2%20%3D%20-0.4055%5C%5Ck%20%3D%200.8110)
So equation is
![T(t) = 40+(30)e^{-0.8110t}\\](https://tex.z-dn.net/?f=T%28t%29%20%3D%2040%2B%2830%29e%5E%7B-0.8110t%7D%5C%5C)
When t=1,
we get
![T(1) = 40+(30)e^{-0.8110}\\\\=53.3324](https://tex.z-dn.net/?f=T%281%29%20%3D%2040%2B%2830%29e%5E%7B-0.8110%7D%5C%5C%5C%5C%3D53.3324)
The gcf of 24 and 48 is 24
Times both sides b 2
2V=(bh)H
remember associative property
(ab)c=a(bc)
and
ab=ba
so
(bh)H=h(bH)
2V=h(bH)
divide both sides by bH
Let
y----------> the length of the rectangle
x---------> the width of the rectangle
P--------> perimeter of a rectangle
we know that
P=38 ft
[P]=2*[x+y]------> 38=2*[x+y]------> equation 1
y=5+x-------> equation 2
<span>substitute 2 in 1
</span>P=2*[x+(5+x)]---------> P=4x+10---------> the given equation
4x=38-10--------> 4x=28----> x=7 ft
y=5+7-----> y=5+7------> y=12 ft
the answer is
the width is 7 ft