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lisov135 [29]
2 years ago
9

Evaluate |3b - 4a| for a = -3 and b = -5.

Mathematics
2 answers:
hram777 [196]2 years ago
7 0

Answer:

3

Step-by-step explanation:

To evaluate |3b - 4a| substitute a = -3 and b = -5 in the given expression.

|3(-5) - 4(-3)| , multiply

|-15 +12| , add

|-3| , take the -3 out of the absolute value

3

Elza [17]2 years ago
3 0

Hey there!

|3b - 4a|

= |3(-5) - 4(-3)|

= |-15 - (-12)|

= |-15 + 12|

= |-3|

= |3|

= 3

Therefore, your answer is: 3

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

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andrey2020 [161]

-5x-9 = -3x -9

Add 3x to both sides:

-2x -9 = -9

Add 9 to both sides:

-2x = 0

Divide both sides by -2:

x = 0

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2 years ago
What would happen if addition were not associative?
bixtya [17]

Answer:

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2 years ago
20y-10x=5 in slope intercept form can you explain the steps
Studentka2010 [4]

Given:

The equation is

20y-10x+5

To find:

The slope intercept form of the given equation.

Solution:

The slope intercept form of a linear equation is

y=mx+b

Where, m is slope and b is the y-intercept.

We have,

20y-10x=5

We need to isolate y. Add 10x on both sides.

20y-10x+10x=5+10x

20y=5+10x

Divide both sides by 20.

\dfrac{20y}{20}=\dfrac{5+10x}{20}

y=\dfrac{5}{20}+\dfrac{10x}{20}

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4 0
3 years ago
Factorise 24e^2-28e-12
Helga [31]

Answer:

4(2e - 3)(3e + 1)

Step-by-step explanation:

Given

24e² - 28e - 12 ← factor out 4 from each term

= 4(6e² - 7e - 3) ← factor the quadratic

Consider the factors of the product of the e² term and the constant term which sum to give the coefficient of the e- term.

product = 6 × - 3 = - 18 and sum = - 7

The factors are - 9 and + 2

Use these factors to split the e- term

6e² - 9e + 2e - 3 ( factor the first/second and third/fourth terms )

= 3e(2e - 3) + 1 (2e - 3) ← factor out (2e - 3) from each term

= (2e - 3)(3e + 1)

Then

24e² - 28e - 12 = 4(2e - 3)(3e + 1) ← in factored form

8 0
3 years ago
~Please help, 50 Points, and Brainliest if correct!~
algol13

A is the correct answer you are looking for

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6 0
3 years ago
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