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3 years ago

14

Carl is boarding a plane.

1 answer:

Reil [10]3 years ago

4 0

if this is an actual question could you give the whole thing?

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If the first and the last terms of an arithmetic series are 10 and 62, show that the sum of the series varies directly as the nu

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Answer:

10+^@=72_^) + the series

Step-by-step explanation:

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3 years ago

PLEASE ANSWER ASAP <br> 20 POINTS

Schach [20]

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3 years ago

BRAINLIEST! :3<br> How much is a $15 tip with the meal costing $18?

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3 years ago

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tamaranim1 [39]

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3 years ago

Read 2 more answers

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