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mojhsa [17]
3 years ago
6

To multiply a monomial by a polynomial, use the Property. Using this property, the product of (2k2 – 7k + 3) and 4k is .

Mathematics
1 answer:
Daniel [21]3 years ago
3 0

Answer:

4k(k-3)(2k-1)

Step-by-step explanation:

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URGENT <br>Perform the indicated operation and simplify the result.
yan [13]

Answer:

The answer to your question is:  (2a - 1) / 4

Step-by-step explanation:

         \frac{12a^{2} - 3}{2} (2a + 1)^{-2}  ( \frac{6}{(2a +1)^{-1} } \\

        \frac{3(4a^{2} -1)}{2} \frac{1}{(2z + 1)^{2} } (\frac{2a + 1)}{6}

        \frac{3(2a + 1)(2a- 1) (2a + 1)}{12(2a + 1)(2a + 1)}

        \frac{(2a - 1)}{4}

8 0
3 years ago
If your car gets 27.6 mi/gal, how many gallons of gasoline would you use if you drove 552.4 miles
valina [46]
The mileage is 27.6 mi/gal
Miles driven = 552.4 mi

Calculate gallons used.
gallons = \frac{552.4 \, mi}{27.6 \, mi/gal} = 20.015 \, gal

Answer: 20.015 gallons  (approximately 20 gallons)

7 0
3 years ago
Read 2 more answers
When graphed, which parabola opens downward? y = –3x2     y = (x – 3)2    y = x2 – 3   
In-s [12.5K]

The answer to this question is y = –3x^2, just had this on e2020 :)

6 0
3 years ago
Read 2 more answers
Help !? please!!!!!!
Lynna [10]

Answer:

slope is 1/3

Step-by-step explanation:

In between the points, you go up one, and forward 3. it can't be a negative slope cause it aint going down. And the slope of 3 is steep last time I checked.

4 0
3 years ago
Read 2 more answers
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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