If p is true and q is false, then p ∨ q is true. A. True B. False

Mathematics

1 answer:

Marysya12 [62]2 years ago

7 0

Answer:

no its false because p v q can only be true if both are true

Step-by-step explanation:

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Naomi has earned $54 mowing lawns the past 2 days. She worded 21/2 hours yesterday and 41/4 hourse today. If Naomi is paid the s

Klio2033 [76]

She worked 2 and 1/2 hours yesterday, and today she did 4 and 1/4 hours.

first off, let's check how many hours she worked total, and divide 54 by it.

so, first off, let's convert the mixed fractions to "improper" and add them up.

$\bf \stackrel{mixed}{2\frac{1}{2}}\implies \cfrac{2\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{5}{2}}
\\\\\\
\stackrel{mixed}{4\frac{1}{4}}\implies \cfrac{4\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{17}{4}}\\\\
-------------------------------$

$\bf \cfrac{5}{2}+\cfrac{17}{4}\impliedby \textit{our LCD will be 4}\implies \cfrac{(2\cdot 5)+(1\cdot 17)}{4}
\\\\\\
\cfrac{10+17}{4}\implies \cfrac{27}{4}\implies \stackrel{total~hours}{6\frac{3}{4}}
\\\\\\
\cfrac{\stackrel{wages}{54}}{\stackrel{hours}{6\frac{3}{4}}}\implies \cfrac{54}{\frac{27}{4}}\implies \cfrac{\frac{54}{1}}{\frac{27}{4}}\implies \cfrac{54}{1}\cdot \cfrac{4}{27}\implies \cfrac{216}{27}\implies 8~\frac{\$}{hr}$

6 0

2 years ago

Match each equation with its number of unique solutions.

Anni [7]

Answer:

y = 3x^2-6x+3 one real solution

y = -x^2 - 4x + 7 two real solution

y = -2x^2+9x-11 two complex solutions

Step-by-step explanation:

b^2-4ac = 0 1 repeated real solution

b^2-4ac > 0 2 distinct real solutions

b^2-4ac < 0 2 complex solutions