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lina2011 [118]
3 years ago
14

Some pls explain this to me ((number 5))

Mathematics
1 answer:
miskamm [114]3 years ago
7 0

Answer:

MP ⊥ LO ⇒ proved

Step-by-step explanation:

In the given figure

∵ m∠LPM = m∠MNO ⇒ given

∵ m∠MNO = m∠MPO ⇒ given

→ If the measure of an angle equals the measures of another two

   angles, then the two angles must be equal in measures

∴ m∠LPM = m∠MPO ⇒ proved

∵ P ∈ line LO

∵ ∠ LPM and ∠MPO are adjacent angles

∴ ∠LPM and ∠MPO formed a pair of linear angles

→ The measure of the linear pair angles is 180°

∴ m∠LPM + m∠MPO = 180°

∵ m∠LPM = m∠MPO

→ That means the measure of each one is 180° divided by 2

∴ m∠LPM = m∠MPO = 180° ÷ 2

∴ m∠LPM = m∠MPO = 90°

∵ MP intersect LO at P and formed right angles at P

→ That means MP is perpendicular to LO

∴ MP ⊥ LO ⇒ proved

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The general form of a solution of the differential equation is already provided for us:

y(x) = c_1 \textrm{e}^x + c_2\textrm{e}^{-x},

where c_1, c_2 \in \mathbb{R}. We now want to find a solution y such that y(-1)=3 and y'(-1)=-3. Therefore, all we need to do is find the constants c_1 and c_2 that satisfy the initial conditions. For the first condition, we have:y(-1)=3 \iff c_1 \textrm{e}^{-1} + c_2 \textrm{e}^{-(-1)} = 3 \iff c_1\textrm{e}^{-1} + c_2\textrm{e} = 3.

For the second condition, we need to find the derivative y' first. In this case, we have:

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This means that the solution obeying the initial conditions is:

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