Answer:
39
Step-by-step explanation:
g(x)= -2x+2 and f(x)= 3x^2+4
(g+f)(-3)
g(-3) = -2(-3) +2 = 6+2 =8
f(-3) = 3 (-3)^2 +4 = 3(9)+4 = 27+4 = 31
(g+f)(-3) = g(-3) + f(-3) = 8+31 = 39
Answer:
⅓y
Step-by-step explanation:
<u>Zeros of the function</u>
f(x) = (x + 2)² - 25
f(x) = (x + 2)(x + 2) - 25
f(x) = x(x + 2) + 2(x + 2) - 25
f(x) = x(x) + x(2) + 2(x) + 2(2) - 25
f(x) = x² + 2x + 2x + 4 - 25
f(x) = x² + 4x + 4 - 25
f(x) = x² + 4x - 21
x² + 4x - 21 = 0
x = <u>-(4) +/- √((4)² - 4(1)(-21))</u>
2(1)
x = <u>-4 +/- √(16 + 84)</u>
2
x = <u>-4 +/- √(100)
</u> 2<u>
</u>x = <u>-4 +/- 10
</u> 2<u>
</u>x = -2 <u>+</u> 5<u>
</u>x = -2 + 5 x = -2 - 5
x = 3 x = -7
f(x) = x² + 4x - 21
f(3) = (3)² + 4(3) - 21
f(3) = 9 + 12 - 21
f(3) = 21 - 21
f(3) = 0
(x, f(x)) = (3, 0)
or
f(x) = x² + 4x - 21
f(-7) = (-7)² + 4(-7) - 21
f(-7) = 49 - 28 - 21
f(-7) = 21 - 21
f(-7) = 0
(x, f(x)) = (-7, 0)
<u>Vertex</u>
<u>X - Intercept</u>
<u />-b/2a = -(4)/2(1) = -4/2 = -2
<u>Y - Intercept</u>
y = x² + 4x - 21
y = (-2)² + 4(-2) - 21
y = 4 - 8 - 21
y = -4 - 21
y = -25
(x, y) = (-2, -25)
<u />
Answer:
0.049168726 light-years
Step-by-step explanation:
The apparent brightness of a star is
where
<em>L = luminosity of the star (related to the Sun)
</em>
<em>d = distance in ly (light-years)
</em>
The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.
Hence the apparent brightness of Alpha Centauri A is
According to the inverse square law for light intensity
where
light intensity at distance 
light intensity at distance 
Let be the distance we would have to place the 50-watt bulb, then replacing in the formula
Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.
