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oksian1 [2.3K]
3 years ago
6

Brian tried to solve a math problem on his math homework. Here's one of the questions he had to answer. How do you solve 2/3p +

3; p = 3/5? Make sure that you show your work!
Mathematics
1 answer:
Bumek [7]3 years ago
8 0

Answer:

The answer would be 17/5

Step-by-step explanation:

To solve this, take the original equation and input 3/5 for p.

2/3p + 3

2/3(3/5) + 3

Now complete the multiplication followed by the addition.

2/3(3/5) + 3

2/5 + 3

17/5

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Write the equation of the line that has a slope of 2/3 and passes through (0,-7).
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The equation of the line is (slope-intercept) y=\frac{2}{3}x+(-7) or (standard form) \frac{-2}{3}x+y=-7
6 0
4 years ago
Use any method to multiply (3a + 2b - c)(a - b + 2c).​
Alla [95]

Answer: 3a^{2}-ab + 5ac- 2b^{2} + 5bc-2c^{2}

Step-by-step explanation:

(3a + 2b − c) (a − b + 2c)  

Expand (3a + 2b − c) (a − b + 2c) by multiplying each  term in the first  expression by each  term in the second expression.

3a ⋅ a + 3a (−b) + 3a (2c) + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)

3 (a ⋅ a) + 3a (−b) + 3a (2c) + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)

3a^2 + 3a (−b) + 3a (2c) + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)

Rewrite using the commutative property of multiplication.

3a^2 + 3 ⋅ −1ab + 3a (2c) + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)

3a^2 − 3ab + 3a (2c) + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)

Rewrite using the commutative property of multiplication.

3a^2 − 3ab + 3 ⋅ 2ac + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)

3a^2 − 3ab + 6ac + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)

Rewrite using the commutative property of multiplication.

3a^2 − 3ab + 6ac + 2ba + 2 ⋅ −1b ⋅ b + 2b (2c) − ca − c (−b) − c (2c)

3a^2 − 3ab + 6ac + 2ba + 2 ⋅ −1 (b ⋅ b) + 2b (2c) − ca − c (−b) − c (2c)

3a^2 − 3ab + 6ac + 2ba + 2 ⋅ −1b^2 + 2b (2c) − ca − c (−b) − c (2c)

3a^2 − 3ab + 6ac + 2ba − 2b^2 + 2b (2c) − ca − c (−b) − c (2c)

Rewrite using the commutative property of multiplication.

3a^2 − 3ab + 6ac + 2ba − 2b^2 + 2 ⋅ 2bc − ca − c (−b) − c (2c)

3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca − c (−b) − c (2c)

Rewrite using the commutative property of multiplication.

3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca − 1 ⋅ −1cb − c (2c)

3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca + 1cb − c (2c)

3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca + cb − c (2c)

Rewrite using the commutative property of multiplication.

3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca + cb − 1 ⋅ 2c ⋅ c

3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca + cb − 1 ⋅ 2 (c ⋅ c)

3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca + cb − 1 ⋅ 2c^2

3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca + cb − 2c^2

3a^2 − 3ab + 2ab + 6ac − 2b^2 + 4bc − ca + cb − 2c^2

3a^2 − ab + 6ac − 2b^2 + 4bc − ca + cb − 2c^2

3a^2 − ab − 2b^2 + 4bc + 6ac − 1ac + cb − 2c^2

3a^2 − ab − 2b^2 + 4bc + 5ac + cb − 2c^2

3a^2 − ab − 2b^2 + 4bc + bc + 5ac − 2c^2

3a^2 − ab − 2b^2 + 5bc + 5ac − 2c^2

3a^2 − ab − 2b^2 + 5ac + 5bc − 2c^2

3a^2 − ab + 5ac − 2b^2 + 5bc − 2c^2

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serg [7]

Answer:

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Step-by-step explanation:

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Makovka662 [10]

Answer:

The first option is the answer (k = +/- 2 square root of J)

Step-by-step explanation:

Since you need to leave k by itself, You multiply both sides by 4 which will leave you with 4J=k^2. To get rid of the square(^2), you need to square root both sides which will leave k by it self. So now you have 4J inside the square root = k. The last step is to take the square root of 4 (which is + or -2) and put it outside the square root sign, leaving J inside the square root.

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Yanka [14]

Answer:

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