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3 years ago

12

Which value of x makes the following equation true? 3(x-2)= -3x 42

1 answer:

Bingel [31]3 years ago

3 0

Well I don't know !
Let's work it out together:

3(x - 2) = -3x + 42

Eliminate parentheses: 3x - 6 = -3x + 42

Add 6 to each side: 3x = -3x + 48

Add 3x to each side: 6x = 48

Divide each side by 6 : x = 8

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What are the 2 ways to make 10's and ones out of 28 cubes

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3 years ago

Please help! Limits. Just confirm that an coverages to 2 as n-->infinity

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3 years ago

Read 2 more answers

Simplify the following

den301095 [7]

Answer:

1) 11 $\sqrt{3}$

2) 2 $\sqrt{2}$

3) $20\sqrt{3} + 15\sqrt{2}$

4) $53 + 12\sqrt{10}$

5) -2

6) $7\sqrt{2} - 5\sqrt{3}$

Step-by-step explanation:

1) 2 $\sqrt{12}$ + 3 $\sqrt{48}$ - $\sqrt{75}$

=(2 × 2 $\sqrt{3}$ )+ (3 × 4 $\sqrt{3}$ ) - 5 $\sqrt{3}$

= 4 $\sqrt{3}$ + 12 $\sqrt{3}$ - 5 $\sqrt{3}$

= 11 $\sqrt{3}$

2) 4 $\sqrt{8}$ -2 $\sqrt{98}$ + $\sqrt{128}$

= (4 × 2 $\sqrt{2}$ ) - (2 × 7 $\sqrt{2}$ ) + 8 $\sqrt{2}$

= 8 $\sqrt{2}$ - 14 $\sqrt{2}$ +8 $\sqrt{2}$

= 2 $\sqrt{2}$

3) 5 $\sqrt{12\\}$ - 3 $\sqrt{18} + 4 \sqrt{72} +2\sqrt{75}$

= 5× $2\sqrt{3}$ - 3× $3\sqrt{2}$ + 4× $6\sqrt{2}$ + 2× $5\sqrt{3}$

= $10\sqrt{3} - 9\sqrt{2} +24\sqrt{2} +10\sqrt{3}$

= $20\sqrt{3} + 15\sqrt{2}$

4) $(2\sqrt{2} + 3\sqrt{5} )^{2}$

= $8 + 12\sqrt{10} + 45$

= $53 + 12\sqrt{10}$

5) $(1+\sqrt{3} ) (1-\sqrt{3} )$

= $1 - 3$

= -2

6) $(2\sqrt{6} -1) (\sqrt{3} -\sqrt{2} )$

= $2\sqrt{18}-2\sqrt{12} -\sqrt{3} +\sqrt{2}$

= 2× $3\sqrt{2}$ - 2× $2\sqrt{3}$ - $\sqrt{3} + \sqrt{2}$

= $6\sqrt{2} - 4\sqrt{3} -\sqrt{3} +\sqrt{2}$

= $7\sqrt{2} - 5\sqrt{3}$

Hope the working out is clear and will help you. :)

5 0

3 years ago

Read 2 more answers

Find two numbers such that one exceeds the other by 11 and their sum is 73.

Whitepunk [10]

Answer: 31 & 42

<u>Step-by-step explanation:</u>

1st #: x

2nd #: x + 11

1st# + 2nd# = sum

x + x + 11 = 73

2x + 11 = 73

2x = 62

x = 31

2nd #: x + 11 = 31 + 11 = 42

5 0

3 years ago