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Oliga [24]
3 years ago
12

Please help problem is in the image.

Mathematics
1 answer:
nika2105 [10]3 years ago
5 0

Answer:

16.5

Step-by-step explanation:

this is beacuse 16.5 times 2 is 33

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1.61.6 248248

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4/5 or 0.8

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What is the correct first step to solve this system of equations by elimination?
storchak [24]

\bold{\huge{\pink{\underline{ Solution }}}}

\bold{\underline{ Given }}

  • <u>We </u><u>have </u><u>given </u><u>two </u><u>linear </u><u>equations </u><u>that</u><u> </u><u>is </u><u>2x </u><u>-</u><u> </u><u>3y </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>and </u><u>x</u><u> </u><u>+</u><u> </u><u>3y </u><u>=</u><u> </u><u>1</u><u>2</u><u> </u><u>.</u>

\bold{\underline{ To \: Find }}

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>x </u><u>and </u><u>y </u><u>by </u><u>elimination </u><u>method</u><u>. </u>

\bold{\underline{ Let's \: Begin }}

\sf{ 2x - 3y = -6 ...eq(1)}

\sf{ x +  3y = 12 ...eq(2)}

<u>Multiply </u><u>eq(</u><u> </u><u>2</u><u> </u><u>)</u><u> </u><u>by </u><u>2</u><u> </u><u>:</u><u>-</u>

\sf{ 2( x + 3y = 12 )}

\sf{ 2x + 6y = 24 }

<u>Subtract </u><u>eq(</u><u>1</u><u>)</u><u> </u><u>from </u><u>eq(</u><u>2</u><u>)</u><u> </u><u>:</u><u>-</u>

\sf{ 2x + 6y -( 2x - 3y) = 24 -(-6)}

\sf{ 2x + 6y - 2x + 3y = 24 + 6 }

\sf{   9y = 30 }

\sf{   y = 30/9}

\sf{\red{ y = 10/3}}

<u>Now</u><u>, </u><u> </u><u>Subsitute</u><u> </u><u>the </u><u>value </u><u>of </u><u>y </u><u>in </u><u>eq(</u><u> </u><u>1</u><u> </u><u>)</u><u>:</u><u>-</u>

\sf{ 2x - 3(10/3) = -6 }

\sf{ 2x - 10 = -6 }

\sf{ 2x  = -6 + 10}

\sf{ x  = 4/2}

\sf{\red{ x  = 2}}

Hence, The value of x and y is 2 and 10/3

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2 years ago
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Step-by-step explanation:

We have the following statement (p\lor q)\lor(\lnot p \land q) \rightarrow q

A truth table shows how the truth or falsity of a compound statement depends on the truth or falsity of the simple statements from which it's constructed.

The simple statements from the statement given are:

\lnot p, p\lor q, \lnot p \land q, and (p\lor q)\lor(\lnot p \land q)

This is the truth table.

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