Factor <img src="https://tex.z-dn.net/?f=x%5E2-2x%2B3" id="TexFormula1" title="x^2-2x+3" alt="x^2-2x+3" align="absmiddle" class=

Does anyone know how to do this?

nekit [7.7K]

That look hard what subject is it

7 0

3 years ago

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Solve for h<br><br> h – 5 = -30<br><br> Make sure to show your work for full points!

Zielflug [23.3K]

H-5 = -30
Add 5 to both sides to get variable h by itself
h = -25

7 0

3 years ago

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A dance studio teaches students a contemporary dance form. It is observed that the probability of students from this studio beco

Helen [10]

$\mathbb P(X=75)=\dbinom{120}{75}0.69^{75}(1-0.69)^{120-75}\approx0.0239$

or about a 2.4% chance.

5 0

3 years ago

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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))

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