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Naddik [55]
3 years ago
5

(On right polynomial) help find equivalent (Left equivalent expressions)

Mathematics
1 answer:
dedylja [7]3 years ago
3 0

Answer:

i

Step-by-step explanation:

i

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Fill in the blank with a constant, so that the resulting expression can be factored as the product of two linear expressions: 2a
Vitek1552 [10]

Answer:

2ab - 6a + 5b - 15

Step-by-step explanation:

Given

2ab - 6a + 5b + \_

Required

Fill in the gap to produce the product of linear expressions

2ab - 6a + 5b + \_

Split to 2

(2ab - 6a) + (5b + \_)

Factorize the first bracket

2a(b - 3) + (5b + \_)

Represent the _ with X

2a(b - 3) + (5b + X)

Factorize the second bracket

2a(b - 3) + 5(b + \frac{X}{5})

To result in a linear expression, then the following condition must be satisfied;

b - 3 = b + \frac{X}{5}

Subtract b from both sides

b - b- 3 = b - b+ \frac{X}{5}

- 3 = \frac{X}{5}

Multiply both sides by 5

- 3 * 5 = \frac{X}{5} * 5

X = -15

Substitute -15 for X in 2a(b - 3) + 5(b + \frac{X}{5})

2a(b - 3) + 5(b + \frac{-15}{5})

2a(b - 3) + 5(b - \frac{15}{5})

2a(b - 3) + 5(b - 3)

(2a + 5)(b - 3)

The two linear expressions are (2a+ 5) and (b - 3)

Their product will result in 2ab - 6a + 5b - 15

<em>Hence, the constant is -15</em>

3 0
3 years ago
a x − y = b a ⁢ x − y = b Which of the following equations are equivalent to the equation above? Select all that apply.
krek1111 [17]
They are all equivalent
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3 years ago
What is m∠A to the nearest degree?
Anuta_ua [19.1K]
So this is a right triangle so you can use sin, cosine, or tangent. for A, you have the sides opposite and hypotenuse, which means you use sin*
sin(A) = opposite / hypotenuse
sin(A) = 56 / 75
arcsin(56/75) = A
A= about 48.3
so to the nearest degree, your answer is 48

*soh cah toa (if you dont know what that is, i can explain it)
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svp [43]

Answer:

Hi I love you and I miss you

I don't know the answear sorry

5 0
2 years ago
What is the answer to this question with work I need it ASAS
PtichkaEL [24]

The answer is 7 3/4 inches.

I noticed the word "train" in the problem and I knew I had to help you! :)

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