Answer:
- The side opposite ∠L is NM.
- The side opposite ∠N is ML.
- The hypotenuse is LN.
Step-by-step explanation:
Even a crude drawing can be helpful as you sort this out. (See attached.)
In general, the side opposite an angle <u>will not</u> have the letter of the angle in its name. Similarly, a side adjacent to the angle <u>will</u> have the angle letter in its name.
The statements that apply are ...
- The side opposite ∠L is NM.
- The side opposite ∠N is ML.
- The hypotenuse is LN. (opposite right angle M)
Answer:
Yeah, D is the right anwser!
Step-by-step explanation:
I would be here for way to long to type up the explanation. And I don't have time to waste.
Answer:
Step-by-step explanation:
Applying the Laplace transform:
![\mathcal{L}[y'']+5\mathcal{L}[y']+4\mathcal{L}[y']=0](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%27%5D%2B5%5Cmathcal%7BL%7D%5By%27%5D%2B4%5Cmathcal%7BL%7D%5By%27%5D%3D0)
With the formulas:
![\mathcal{L}[y'']=s^2\mathcal{L}[y]-y(0)s-y'(0)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%27%5D%3Ds%5E2%5Cmathcal%7BL%7D%5By%5D-y%280%29s-y%27%280%29)
![\mathcal{L}[y']=s\mathcal{L}[y]-y(0)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%5D%3Ds%5Cmathcal%7BL%7D%5By%5D-y%280%29)
![\mathcal{L}[x]=L](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5Bx%5D%3DL)
Solving for 
Apply the inverse Laplace transform with this formula:
![\mathcal{L}^{-1}[\frac1{s-a}]=e^{at}](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5E%7B-1%7D%5B%5Cfrac1%7Bs-a%7D%5D%3De%5E%7Bat%7D)
![y=3\mathcal{L}^{-1}[\frac1{s+4}]=3e^{-4t}](https://tex.z-dn.net/?f=y%3D3%5Cmathcal%7BL%7D%5E%7B-1%7D%5B%5Cfrac1%7Bs%2B4%7D%5D%3D3e%5E%7B-4t%7D)
The answer is
C) -7
Multiply -5 by 2
Divide 8 by 4 :
3-10+2-2
Subtract 10 from 3 :
-7+2-2
Add -7 and 2 :
-5 - 2
Subtract 2 from -5:
-7
Answer:
(- 1, 1 )
Step-by-step explanation:
Given the 2 equations
y = 3x + 4 → (1)
y = x + 2 → (2)
Substitute y = 3x + 4 into (2)
3x + 4 = x + 2 ( subtract x from both sides )
2x + 4 = 2 ( subtract 4 from both sides )
2x = - 2 ( divide both sides by 2 )
x = - 1
Substitute x = - 1 into either of the 2 equations and evaluate for y
Substituting x = - 1 into (2)
y = - 1 + 2 = 1
Solution is (- 1, 1 )
