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3 years ago

When the following expression is written in simplest form, what is the coefficient of the variable term? -2.5y + 8y - 10.5y

Mathematics

1 answer:

LUCKY_DIMON [66]3 years ago

7 0

Answer:

Option 1 is correct.

-5

Step-by-step explanation:

Given:

The given expression is.

$=-2.5y+8y-10.5y$

Write the given expression in simplest form.

$=-2.5y+8y-10.5y$

Add negative term of y.

$=-13y+8y$

$=-5y$

The simplest form of given expression is $=-5y$

Therefore, the coefficient of the variable term is -5

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How can I factor 5x^2 + 26x - 24 = 0 using the completing the square method.

kifflom [539]

<h2><u>QUADRATIC EQUATION</u></h2>

<h2> $\mathbb{ANSWER:}$ </h2>

$\bold{x = 0.8 \: } \: \sf \: {\color{grey}or} \: \: \: \bold{ x= \frac{4}{5} } \\$

$\bold{x = - 6}$

— — — — — — — — — —

<h3>Step-by-step explanation:</h3>

<u>How can </u><u>we</u><u> factor 5x^2 + 26x - 24 = 0 using the completing the square </u><u>method?</u>

Let's solve your equation step-by-step.

$\bold{Given \: Equation: \color{brown} 5x²+26x-24=0}$

First, add 24 to both sides.

$\bold{5x²+26x-24 - \purple{ 24} = 0 + \purple{24}}$

$\bold{ \implies \: 5x²+26x = 24 }$

Since the coefficient of 5x² is 5, divide both sides by 5.

$\bold{ \frac{5 {x}^{2} + 26x }{5} = \frac{24}{5} } \\$

$\bold{ \implies \: {x}^{2} + \frac{26}{5} x = \frac{24}{5} } \\$

The coefficient of 26/5x is 26/5. So, let b=26/5.

Then we need to add (b/2)²=169/25 to both sides to complete the square.

Add 169/25 to both sides.

$\bold{ {x}^{2} + \frac{26}{5} x + \frac{ \purple{169}}{ \purple{25}}= \frac{24}{5} + \frac{ \purple{169}}{ \purple{25}} } \\$

$\bold{ \implies \:\bold{ {x}^{2} + \frac{26}{5} x + \frac{ 169}{ {25}}= \frac{289}{25} } } \\$

Factor the left side.

$\bold{(x + \frac{13}{5} ) {}^{2} = \frac{289}{25} } \\$

Take square root.

$\bold{x + \frac{13}{5} = ± \: \sqrt{ \frac{289}{25} }} \\$

Then, add (-13)/5 to both sides.

$\bold{x + \frac{13}{5} + \frac{\purple{ - 13} }{\purple{ 5}} = \frac{{ - 13} }{{ 5}} ± \: \sqrt{ \frac{289}{25}}} \\$

$\bold{ \: x = \frac{ - 13}{5} ± \sqrt{ \frac{289}{25} } } \\$

$\implies \: \underline{ \boxed{ \bold{ \frac{4}{5} } \sf \: \: or \: \: \bold{ x = - 6}}} \\$

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