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3 years ago

Write the equation x² + 6x + y² – 8y = 19 in standard form.

Mathematics

2 answers:

kirza4 [7]3 years ago

8 0

Answer:

Formula used :

x²+2xy+y² = (x+y)²
x²-2xy+y² = (x-y)²

$→{x}^{2} + 6x + {y}^{2} - 8y = 19 \\ ({x}^{2} + 2.3.x + {3}^{2} ) +({y}^{2} - 2.4.y + {4}^{2} ) = 19 + {3}^{2} + {4}^{2} \\ {(x + 3)}^{2} + {(y - 4)}^{2} = 19 + 9 + 16 \\ \boxed{ {(x + 3)}^{2} + {(y - 4)}^{2} = 44}✓$

3) (x+3)²+(y-4)²=44 is the right answer.

Rama09 [41]3 years ago

4 0

Answer:

option C

Step-by-step explanation:

using (X + 3)² + ( y - 4)² = 44

gives you directly x² + 6x + y² - 8y = 19

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2 years ago

Write the equation of the line perpendicular to 2x+3y=9 that passes through (-2,5). Write your answer in slope-intercept form. S

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ANSWER

$y = \frac{3}{2} x + 8$

EXPLANATION

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$2x + 3y = 9$

We need to write this equation in the slope intercept form to obtain,

$3y = - 2x + 9$

$\Rightarrow \: y = - \frac{2}{3}x + 3$

The slope of this line is

$m_1 = - \frac{2}{3}$
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$m_2$

Then
$m_1 \times m_2 = - 1$

$- \frac{2}{3} m_2= - 1$

This implies that,

$m_2 = - 1 \times - \frac{3}{2}$

$m_2 = \frac{3}{2}$

Let the equation of the perpendicular line be,

$y = mx + b$

We substitute the slope to get,

$y = \frac{3}{2} x + b$

Since this line passes through
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This means that,

$5= \frac{3}{2} ( - 2)+ b$

$5 = - 3 + b$

$5 + 3 = b$

$b = 8$

Wherefore the slope-intercept form is

$y = \frac{3}{2} x + 8$

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4 years ago

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Remember the power rule for differentiation shown below:

$\frac{d}{dx}(x^n)=nx^{n-1}$