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2 years ago

33. Construct a truth table to show that the following argument is valid.

Mathematics

1 answer:

GuDViN [60]2 years ago

5 0

Answer:

see attached

Step-by-step explanation:

Perhaps you want to show that P∨Q will be equal to Q when ¬P is true. Attached is a truth table that shows this is the case.

This is simply a restatement of the logic of the OR function. When one its inputs is a logical 0 (false), the output depends on the other input(s).

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3 years ago

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How many solutions does the system of equations below have?

soldier1979 [14.2K]

Answer:

One solution

Step-by-step explanation:

5x + y = 8

15x + 15y = 14

Lets solve using substitution, first we need to turn "5x + = 8" into "y = mx + b" or slope - intercept form

So we solve for "y" in the equation "5x + y = 8"

5x + y = 8

Step 1: Subtract 5x from both sides.

5x + y − 5x = 8 − 5x

Step 2: 5x subtracted by 5x cancel out and "8 - 5x" are flipped

y = −5x + 8

Now we can solve using substitution:

We substitute "-5x + 8" into the equation "15x + 15y = 14" for y

So it would look like this:

15x + 15(-5x + 8) = 14

Now we just solve for x

15x + (15)(−5x) + (15)(8) = 14(Distribute)

15x − 75x + 120 = 14

(15x − 75x) + (120) = 14(Combine Like Terms)

−60x + 120 = 14

Step 2: Subtract 120 from both sides.

−60x + 120 − 120 = 14 − 120

−60x = −106

Divide both sides by -60

$\dfrac{ -60x }{ -60 } = \dfrac{ -106 }{ -60 }$

Simplify

$x = \dfrac{ 53 }{ 30 }$

Now that we know the value of x, we can solve for y in any of the equations, but let's use the equation "y = −5x + 8"

$\mathrm{So\:it\:would\:look\:like\:this:\ y = -5 \left( \dfrac{ 53 }{ 30 } \right) +8}$

$\mathrm{Now\:lets\:solve\:for\:"y"\:then}$

$y = -5 \left( \dfrac{ 53 }{ 30 } \right) +8}$

$\mathrm{Express\: -5 \times \dfrac{ 53 }{ 30 }\:as\:a\:single\:fraction}$

$y = \dfrac{ -5 \times 53 }{ 30 } +8$

$\mathrm{Multiply\:-5 \:and\:53\:to\:get\:-265 }$

$y = \dfrac{ -265 }{ 30 } +8$

$\mathrm{Simplify\: \dfrac{ -265 }{ 30 } \:,by\:dividing\:both\:-265\:and\:30\:by\:5} }$

$y = \dfrac{ -265 \div 5 }{ 30 \div 5 } +8$

$\mathrm{Simplify}$

$y = - \dfrac{ 53 }{ 6 } +8$

$\mathrm{Turn\:8\:into\:a\:fraction\:that\:has\:the\:same\:denominator\:as\: - \dfrac{ 53 }{ 6 }}$

$\mathrm{Multiples\:of\:1: \:1,2,3,4,5,6}$

$\mathrm{Multiples\:of\:6: \:6,12,18,24,30,36,42,48}$

$\mathrm{Convert\:8\:to\:fraction\:\dfrac{ 48 }{ 6 }}$

$y = - \dfrac{ 53 }{ 6 } + \dfrac{ 48 }{ 6 }$

$\mathrm{Since\: - \dfrac{ 53 }{ 6 }\:have\:the\:same\:denominator\:,\:add\:them\:by\:adding\:their\:numerators}$

$y = \dfrac{ -53+48 }{ 6 }$

$\mathrm{Add\: -53 \: and\: 48\: to\: get\: -5}$

$y = - \dfrac{ 5 }{ 6 }$

$\mathrm{The\:solution\:is\:the\:ordered\:pair\:(\dfrac{ 53 }{ 30 }, - \dfrac{ 5 }{ 6 })}$

So there is only one solution to the equation.

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