If (a-2)^2+(b+1)^2+(c-3)2 =0 then the value of a^2/4+b^2+c^2/3

Mathematics

2 answers:

almond37 [142]3 years ago

7 0

The correct answer is 2. a = 1/2, b = - 3, c = -2

Step-by-step explanation:

1. Let's review the information provided for an answering the question correctly:

0 =1/2x^2- 3x- 2

0 = 1/2x² - 3x - 2

2. Let's find the values for a, b and c.

According to the quadratic equation, we have:

a = 1/2

b = - 3

c = - 2

The correct answer is 2. a = 1/2, b = - 3, c = -2

(❤´艸｀❤)

aliya0001 [1]3 years ago

6 0

Answer:

Step-by-step explanation:

(a-2)^2+(b+1)^2+(c-3)^2 =0

Since all 3 terms are square numbers, they must be = 0 to have a sum of 0.

so a-2= 0, b+1=0 and c-3= 0

a=2, b= -1, c= 3

a^2/4+b^2+c^2/3= 4/4+1+9/3= 1+1+3= 5

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Given that f.x 3x-2 over x+1 g[x] x +5 evaluate f[-4] and gf [-2]

Jobisdone [24]

The value of f[ -4 ] and g°f[-2] are $\frac{14}{3}$ and 13 respectively.

<h3>What is the value of f[-4] and g°f[-2]?</h3>

Given the function;

$f(x) = \frac{3x-2}{x+1}$
$g(x)=x+5$
f[ -4 ] = ?
g°f[ -2 ] = ?

For f[ -4 ], we substitute -4 for every variable x in the function.

$f(x) = \frac{3x-2}{x+1}\\\\f(-4) = \frac{3(-4)-2}{(-4)+1}\\\\f(-4) = \frac{-12-2}{-4+1}\\\\f(-4) = \frac{-14}{-3}\\\\f(-4) = \frac{14}{3}$

For g°f[-2]

g°f[-2] is expressed as g(f(-2))

$g(\frac{3x-2}{x+1}) = (\frac{3x-2}{x+1}) + 5\\\\g(\frac{3x-2}{x+1}) = \frac{3x-2}{x+1} + \frac{5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) = \frac{3x-2+5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) = \frac{8x+3}{x+1}\\\\We\ substitute \ in \ [-2] \\\\g(\frac{3x-2}{x+1}) = \frac{8(-2)+3}{(-2)+1}\\\\g(\frac{3x-2}{x+1}) = \frac{-16+3}{-2+1}\\\\g(\frac{3x-2}{x+1}) = \frac{-13}{-1}\\\\g(\frac{3x-2}{x+1}) = 13$