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tankabanditka [31]
3 years ago
11

10( 7+6s) its distributing

Mathematics
2 answers:
Fofino [41]3 years ago
4 0

Answer:

10(7 + 6s) = 70 + 60s

Step-by-step explanation:

10(7 + 6s)

Distribute the 10 in the parenthesis. When distributing, multiply the number by every term in the parenthesis.

10 * 7 + 10 * 6s

Multiply.

70 + 60s

algol [13]3 years ago
4 0

Answer:

10(7)+10(6s) which equals 70+60s

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In a quadrilateral ABCD, angle A + angle D = 90°. Prove that AC^2 + BD^2 = BC^2 + AD^2​
olga2289 [7]

Answer:

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

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

Hence prove.

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

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°

∠E + (∠A + ∠D) = 180°

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

∠E + 90° = 180°

∠E = 180° - 90°

∠E = 90°

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}

BD^{2}=BE^{2}+ED^{2}

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}

4 0
3 years ago
Please help me I don't understand
zalisa [80]
The answer is d you would divide both sides by pi to get d
5 0
3 years ago
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