Answer:
Step-by-step explanation:
The opposite side (the one not connected to A) = 4
The hypotenuse is 5
The adjacent side needs to be found for the cosine and the tangent.
a^2 + b^2 = c^2
a = opposite side = 4
b = adjacent side = ?
c = hypotenuse = 5
4^2 + x^2 = 5^2
16 + x^2 = 25
x^2 = 25 - 16
x^2 = 9
x = sqrt(9)
x = 3
cos(A) = adjacent / hypotenuse = 3/5
Tan(A) = opposite / adjacent = 4/3
cos(A) + tan(A) = 3/5 + 4/3
cos(A) + tan(A) = 9/15 + 20/15 = 29/15
Answer:
1.214 or 
or 
Step-by-step explanation:
To solve this equation, first you need to the variable term isolated. You do this by adding 10 to each side. Your equation will now look like
17c = 14
Next, you will divide both sides by the 17.
Your equation now is solved, and above, I put ther three different forms for your answer: decimal, mixed number, and improper fraction.
~theLocoCoco
Answer:
a) W₁ = 78400 [J]
b)Wt = 82320 [J]
Step-by-step explanation:
a) W = ∫ f*dl general expression for work
If we have a chain with density of 10 Kg/m, distributed weight would be
9.8 m/s² * 10 kg = mg
Total length of th chain is 40 m, and the function of y at any time is
f(y) = (40 - y ) mg where ( 40 - y ) is te length of chain to be winded
At the beggining we have to wind 40 meters y = 0 at the end of the proccess y = 40 and there is nothing to wind then:
f(y) = mg* (40 - y )
W₁ = ∫f(y) * dy ⇒ W₁ = ∫₀⁴⁰ mg* (40 - y ) dy ⇒ W₁ = mg [ ∫₀⁴⁰ 40dy - ∫₀⁴⁰ ydy
W₁ = mg [ 40*y |₀⁴⁰ - 1/2 * y² |₀⁴⁰ ⇒ W₁ = mg* [ 40*40 - 1/2 (40)² ]
W₁ = mg * [1/2] W₁ = 10*9,8* ( 800 )
W₁ = 78400 [J]
b) Now we can calculate work to do if we have a 25 block and the chain is weightless
W₂ = ∫ mg* dy ⇒ W₂ = ∫₀⁴⁰ mg*dy ⇒ W₂ = mg y |₀⁴⁰
W₂ = mg* 40 = 10*9.8* 40
W₂ = 3920 [J]
Total work
Wt = W₁ + W₂ ⇒ Wt = 78400 + 3920
Wt = 82320 [J]
Answer:
5
Step-by-step explanation:
Answer:
-15
Step-by-step explanation:
