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Naddika [18.5K]
2 years ago
9

Which of the following is equivalent to 30p4q3r + 42p2r - 18p2q2r when it is completely factored?

Mathematics
2 answers:
Eddi Din [679]2 years ago
7 0
<span>6(p^2)r[5(p^2)(q^3) + 7 - 3(q^2)]  </span>while the GCFs are 6, p^2, and r
Hope this Helps (:
 
solmaris [256]2 years ago
7 0

Answer:

The answer is D

Step-by-step explanation:

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PLEASE HELP ANSWER ASAP I WILL GIVE BRAINLIEST
AysviL [449]

Answer:

Let number of weeks be x

10x = 200-150

x=50/10

x=5

Thus it would take him 5 weeks to buy himself a iPhone12

Hope it helps

8 0
3 years ago
The game of Connex contains one 4-unit piece, two identical 3-unit pieces, three identical 2-unit pieces and four identical 1-un
Aleks [24]

Answer: 277 ways

Step-by-step explanation:

Let’s start bycreating 10-unit pieces using the 4-unit piece.

The arrangements are:

1). 4-3-3 (3 permutations)

2). 4-3-2-1 = 4! = 24 permutations.

3). 4-3-1-1-1 (5*[4!/(3!1!)]

= 5*4

= 20permutations

4). 4-2-2-2 (4 permutations)

5). 4-2-2-1-1 (5 *[4!/(2!2!)]

= 5*6

= 30 permutations

6). 4-2-1-1-1-1(6*[5!/(4!1!)]

= 6*5

= 30 permutations

Let’s-consider the arrangements using one or more3-unit pieces and no 4-unit piece:

7). 3-3-2-2 (4!/(2!2!)

8). 3-3-2-1-1 (5*(4!/(2!2!)

= 5*6

= 30 permutations.

9). 3-3-1-1-1-1 (6!/(4!2!) = 0

10). 3-2-2-2-1 (5*4!/(3!1!)

= 5*4

= 20permutations

11). 3-2-2-1-1-1 (6*5!/(3!2!)

= 6*10

= 60 permutations

Finally we would look at the arrangements using only 1-and 2-unit pieces:

12). 2-2-2-1-1-1-1 (7!/(4!3!)

= 35 permutations

add them all up:

(3 + 24 + 20 + 4) + (30 + 30 + 6 + 30) + (15 + 20 + 60 + 35)

=51 + 96 + 130

= 277ways

5 0
2 years ago
What is the correct answer and how can this be solved?
pav-90 [236]

Answer:

$\mathbf{\frac{1}{19} }

Step-by-step explanation:

$$\bullet \Nth \ Term;\\$$$\frac{n+2}{2n^{2} +3n-2}

$$\bullet U_{10} \ Term;\\\\$$\boxed{\frac{(10+2) }{2*10^{2} +3*10-2}=  \frac{1}{19} }

5 0
3 years ago
Read 2 more answers
What is the difference of the rational expressions below? x/x-2 - 3/x
Lera25 [3.4K]

Answer:

x^2-3x+6/x^2-2x

Step-by-step explanation:

5 0
3 years ago
What are the real and complex solutions of the polynomial equation X^3-216=0
vazorg [7]

Answer:

6,\ -3+3\sqrt{3}i,\ -3-3\sqrt{3}i

Step-by-step explanation:

Factor the equation x^3-216=0 using formula for difference of the cubes:

x^3-216=(x-6)(x^2+6x+36),

then

x-6=0\ or\ x^2+6x+36=0.

1. The equation x-6=0 has real solution x=6.

2. The equation x^2+6x+36=0 has negative discriminant D=6^2-4\cdot 36=36-144=-108, then it has two complex solutions

x_{1,2}=\dfrac{-6\pm 6\sqrt{3}i}{2}=-3\pm3\sqrt{3}i.

3 0
3 years ago
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