Answer:
AC and DF
Step-by-step explanation:
We can see that the triangles have a scale factor of 3.
1 * 3 = 3 (AB and DE)
2 * 3 = 6 (BC and EF)
However...
2 * 3 = 6 not 7 (AC and DF)
Hello there!
(6y^2 - 3y^3 + 5)(7y^5)
Use the distributive property
=(6y^2)(7y^5) + (-3y^3)(7y^5) + (5)(7y^5)
= 42y^7 - 21y^8 + 35y^5
= -21y^8 + 42y^7 + 35y^5
Good luck with your studies buddy!
Given that
Sin θ = a/b
LHS = Sec θ + Tan θ
⇛(1/Cos θ) + (Sin θ/ Cos θ)
⇛(1+Sin θ)/Cos θ
We know that
Sin² A + Cos² A = 1
⇛Cos² A = 1-Sin² A
⇛Cos A =√(1-Sin² A)
LHS = (1+Sin θ)/√(1- Sin² θ)
⇛ LHS = {1+(a/b)}/√{1-(a/b)²}
= {(b+a)/b}/√(1-(a²/b²))
= {(b+a)/b}/√{(b²-a²)/b²}
= {(b+a)/b}/√{(b²-a²)/b}
= (b+a)/√(b²-a²)
= √{(b+a)(b+a)/(b²-a²)}
⇛ LHS = √{(b+a)(b+a)/(b+a)(b-a)}
Now, (x+y)(x-y) = x²-y²
Where ,
On cancelling (b+a) then
⇛LHS = √{(b+a)/(b-a)}
⇛RHS
⇛ LHS = RHS
Sec θ + Tan θ = √{(b+a)/(b-a)}
Hence, Proved.
<u>Answer</u><u>:</u> If Sinθ=a/b then Secθ+Tanθ=√{(b+a)/(b-a)}.
<u>also</u><u> read</u><u> similar</u><u> questions</u><u>:</u><u>-</u> i) sin^2 A sec^2 B + tan^2 B cos^2 A = sin^2A + tan²B..
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Sec x -tan x sin x =1/secx Help me prove it..
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Answer:
$102.50
Step-by-step explanation:
The formula here would be x=35+15y, y being the number of hours he was there.
So 35+15(4.5)=102.5, or $102.50.
Answer:
5j
Step-by-step explanation:
