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Zanzabum
3 years ago
7

-7m+10<-39 Solve the Inequality using the format M<1

Mathematics
1 answer:
love history [14]3 years ago
8 0

Answer:

m>7

Step-by-step explanation:

-7m+10<-39

-7m<-39-10

-7m<-49

7m>49

m>7

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In a quadrilateral ABCD, angle A + angle D = 90°. Prove that AC^2 + BD^2 = BC^2 + AD^2​
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Answer:

AD^{2}+BC^{2}=AE^{2}+ED^{2}+BE^{2}+EC^{2}

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Step-by-step explanation:

Given:

∠A + ∠D = 90°

We are prove to that

AC^{2}+BD^{2}=BC^{2}+AD^{2}

Solution:

See required figure in attached file.

We know sum of the all angles of a triangle is 180°.

So, in triangle AED.

∠A + ∠E + ∠D = 180°

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Now, we substitute ∠A + ∠D = 90° in above equation.

∠E + 90° = 180°

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Using Pythagoras Theorem for triangle ADE and triangle BEC.

AD^{2}=AE^{2}+ED^{2}

BC^{2}=BE^{2}+EC^{2}

Now, we add both above equations.

AD^{2}+BC^{2}=AE^{2}+ED^{2}+BE^{2}+EC^{2}--------(1)

Similarly, Using Pythagoras Theorem for triangle AEC and triangle BED.

AC^{2}=AE^{2}+EC^{2}

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Now, we add both above equations.

AC^{2}+BD^{2}=AE^{2}+EC^{2}+BE^{2}+ED^{2}--------(2)

We get From equation 1 and equation 2.

AC^{2}+BD^{2}=BC^{2}+AD^{2}

Hence prove,

AC^{2}+BD^{2}=BC^{2}+AD^{2}

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{s | s <90}

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