2(5c+ 2) - 2c = 3(2c + 3) + 7

Mathematics

2 answers:

Mademuasel [1]2 years ago

8 0

Answer:

c = 6

Step-by-step explanation:

Step 1: Write equation

2(5c + 2) - 2c = 3(2c + 3) + 7

Step 2: Solve for <em>c</em>

Distribute: 10c + 4 - 2c = 6c + 9 + 7
Combine like terms: 8c + 4 = 6c + 16
Subtract 6c on both sides: 2c + 4 = 16
Subtract 4 on both sides: 2c = 12
Divide both sides by 2: c = 6

bearhunter [10]2 years ago

3 0

The answer for this is c = 6

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The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
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$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
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Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
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$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!