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user100 [1]
3 years ago
15

Direction: Simplify each expression using Product Rule.

Mathematics
1 answer:
Sever21 [200]3 years ago
3 0

Answer:

pagal answer toone khoo

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Shawna has 4/5 ounces of juice. She drinks 3/4 of it. How much is left
Bezzdna [24]

Answer:

0.05

Step-by-step explanation:

I turned the fractions into decimals. to do that your do exactly as it tells you!

4 divided by 5 = 0.8

so she has 0.8 ounces of juice (That not alot lol I would not be satisfied) now for how much she drank...

3 divided by 4 =0.75

0.80 - 0.75 = 0.05

She drank 0.05 ounces

Hope this helped! Please mark as brainliest! Thanks! If you need me to turn 0.05 in a fraction i can do that.

4 0
3 years ago
Show work please I need HELP QUICK PLEASE
Genrish500 [490]

Answer:

Step-by-step explanation:

perimeter = sum of the sides = (3y+5) + (y-4) + (6y) = 10y + 1

7 0
3 years ago
I will give you brainliest whoever answers correct first!!
Firlakuza [10]

Answer:

74.4 (llllllllllll ignore this i need 20 characters to answer)

3 0
3 years ago
What is the volume of the cylinder below if radius is 11 and height is 15?
almond37 [142]

volume of cylinder = PI*r^2*h

 r=11

h=15

 using 3.14 for pi

3.14*11^2*15 = 5699.1 units^3

3 0
3 years ago
Recall, we have five connectives in propositional logic ¬, ∧, ∨, →, ↔ (negation, conjunction, disjunction, conditional and bicon
Free_Kalibri [48]

Answer:

(a) ¬(p→¬q)

(b) ¬p→q

(c) ¬((p→q)→¬(q→p))

Step-by-step explanation

taking into account the truth table for the conditional connective:

<u>p | q | p→q </u>

T | T |   T    

T | F |   F    

F | T |   T    

F | F |   T    

(a) and (b) can be seen from truth tables:

for (a) <u>p∧q</u>:

<u>p | q | ¬q | p→¬q | ¬(p→¬q) | p∧q</u>

T | T |  F  |   F     |    T       |  T

T | F |  T  |  T      |    F       |  F

F | T |  F  |  T      |    F       |  F

F | F |  T  |  T      |    F       |  F

As they have the same truth table, they are equivalent.

In a similar manner, for (b) p∨q:

<u>p | q | ¬p | ¬p→q | p∨q</u>

T | T |  F  |   T     |    T    

T | F |  F  |   T     |    T    

F | T |  T  |   T     |    T    

F | F |  T  |   F     |    F    

again, the truth tables are the same.

For (c)p↔q, we have to remember that p ↔ q can be written as (p→q)∧(q→p). By replacing p with (p→q) and q with (q→p) in the answer for part (a) we can change the ∧ connector to an equivalent using ¬ and →. Doing this we get ¬((p→q)→¬(q→p))

4 0
3 years ago
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