Answer:
Step-by-step explanation:
So, the longest tooth is 4 3/4 and the shortest is 3 1/4. So if you subtract 4 3/4 - 3 1/4, you will get the answer 1 2/4, which you can simplify to 1 1/2 or 1.5 :)
The sides are (TQ and HE), (RS and GF), (HG and ST), (RQ and EF)
Answer:
I think it is a scalene acute because:
➡It has acute angles
➡Its angles are not equal ⬇
Scalene triangles do not have equal angles
Hope this helps, adn have a good day! ฅ^•ﻌ•^ฅ
Let r = (t,t^2,t^3)
Then r' = (1, 2t, 3t^2)
General Line integral is:
![\int_a^b f(r) |r'| dt](https://tex.z-dn.net/?f=%5Cint_a%5Eb%20f%28r%29%20%7Cr%27%7C%20dt)
The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector
![\sqrt{(x')^2 + (y')^2 + (z')^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28x%27%29%5E2%20%2B%20%28y%27%29%5E2%20%2B%20%28z%27%29%5E2%7D)
![\int_0^1 (2t+9t^3) \sqrt{1+4t^2 +9t^4} dt](https://tex.z-dn.net/?f=%5Cint_0%5E1%20%282t%2B9t%5E3%29%20%5Csqrt%7B1%2B4t%5E2%20%2B9t%5E4%7D%20dt%20)
Fortunately, this simplifies nicely with a 'u' substitution.
Let u = 1+4t^2 +9t^4
du = 8t + 36t^3 dt
![\int_0^1 \frac{2t+9t^3}{8t+36t^3} \sqrt{u} du \\ \\ \int_0^1 \frac{2t+9t^3}{4(2t+9t^3)} \sqrt{u} du \\ \\ \frac{1}{4} \int_0^1 \sqrt{u} du](https://tex.z-dn.net/?f=%5Cint_0%5E1%20%5Cfrac%7B2t%2B9t%5E3%7D%7B8t%2B36t%5E3%7D%20%5Csqrt%7Bu%7D%20%20du%20%5C%5C%20%20%5C%5C%20%5Cint_0%5E1%20%5Cfrac%7B2t%2B9t%5E3%7D%7B4%282t%2B9t%5E3%29%7D%20%5Csqrt%7Bu%7D%20%20du%20%5C%5C%20%20%5C%5C%20%20%5Cfrac%7B1%7D%7B4%7D%20%5Cint_0%5E1%20%5Csqrt%7Bu%7D%20%20du)
After integrating using power rule, replace 'u' with function for 't' and evaluate limits:
Let l the length of a side edge and d the length of diagonal of a side
so than using sin we can writing
sin45 = l/d
sqrt2 /2 = l/d
and given that this area s=5sqrt2 what is equal l*d so in this case there are
l/d = sqrt2 /2
l*d = 5sqrt2 => l = 5sqrt2 /d and this subtituted in place of l inside the first equation we get
(5sqrt2 /d)/d = sqrt2 /2
5sqrt2 /d^2 = sqrt2 /2
10sqrt2 = d^2 *sqrt2 divide both sides by sqrt2
10 = d^2
d = sqrt10 so than d = sqrt10 result l = 5sqrt2 /sqrt10 = 5sqrt20 /10 =
= sqrt20 /2 = 2sqrt5 /2 = sqrt5
l = sqrt5
d = sqrt10
so area total del cubo = 6*l^2 = 6*(sqrt5)^2 = 6*5 = 30 unit squared
so choice d) is right sure
hope this will help you