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xeze [42]
3 years ago
12

Let f be a differentiable function such that f(3) = 2.345 and f' (x) = ln (x^2+ 1). What is the value of f (5)?

Mathematics
1 answer:
Rudik [331]3 years ago
5 0

Use a linear approximation to f(x) centered at x=3:

f(x)\approx f(3) + f'(3)(x-3)

Then

f(5)\approx2.345+\ln(3^2+1)(5-3)\approx6.95

which makes C the most likely choice.

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yz = 7ln(x + z), (0, 0, 1) (a) the tangent plane Correct: Your answer is correct. (b) parametric equations of the normal line to
Alika [10]

Answer:

The equation of tangent plane is    7x - y + 7z - 7 = 0

Parametric equation of normal line

x = 7 t         , y=-t                 , z=1+7 t

Step-by-step explanation:

Equation of tangent

fₓ (x₀ , y₀ , z₀) (x-x₀) + fy (x₀ , y₀ , z₀) (y-y₀) +fz(x₀ , y₀ , z₀)(z-z₀)=0                (1)

From taking derivation we get

fₓ (x₀ , y₀ , z₀) = 7

fy (x₀ , y₀ , z₀)= -1

fz(x₀ , y₀ , z₀) = 7

putting these value in equation 1

 (7) (x-0) + (-1)(y-0) + 7(z-1)=0

  7x - y + 7z - 7 = 0

The equation of tangent plane is    7x - y + 7z - 7 = 0

b) Parametric equation

x=x +f ₓ (P)t , y = y₀ +f y (P) t , z=z₀ +f z (P) t

x=0 +7 t      , y =0+(-1) t        , z=1+7 t

x=7 t            ,y=-t                  , z= 1+7 t

Parametric equation of normal line

x = 7 t         , y=-t                 , z=1+7 t

7 0
4 years ago
HElp please pleassssssssss i need help E E E HEEEELLLLLLPPPPP!!!!!!11
Romashka-Z-Leto [24]

Answer:

6.3

Step-by-step explanation:

square root of 40 is 6.32455532034

rounded to nearest tenth is 6.3

3 0
4 years ago
Read 2 more answers
Noah bought 15 baseball cards for $9. At this rate, what is the cost of 1 baseball card?
pychu [463]
You do nine divided by fifteen which =$0.6
3 0
3 years ago
Read 2 more answers
1. Determinar la ecuación canónica de la parábola con vértice en (-2,4) y foco en (1,4) 2. Determinar el foco y el vértice de la
goldenfox [79]

Answer:

1. La ecuación de la parábola en forma canónica es x = 1/12 × (y - 4) ² - 2

2. Vértice = (-1, 3), enfoque = (-5/2, 3)

3. y = 12x no es una parábola

4. y = -8x, no es una parábola

Step-by-step explanation:

1. La ecuación estándar de una parábola es y = a · x² + b · x + c

El vértice V es (h, k)

El foco (h + p, k)

Por lo tanto, tenemos en comparación k = 4, h = -2

h + p = 1

p = 1 - h = 1 - (-2) = 3

Lo que da la ecuación como (y - k) ² = 4 · p · (x - h)

Al ingresar los valores de k, h y p, tenemos

(y - 4) ² = 4 × 3 × (x - (-2)) = 12 × (x + 2)

12 · x + 24 = (y - 4) ²

x = 1/12 × (y - 4) ² - 2

La ecuación de la parábola en forma canónica es x = 1/12 × (y - 4) ² - 2

2. Determinar el foco y el vértice de la parábola (y - 3) ² = -6 · (x + 1)

Reescribimos la ecuación en forma de vértice de la siguiente manera;

-6 · x -6 = (y - 3) ²

x = -1 / 6 × (y - 3) ² - 1

La ecuación de una parábola en forma de vértice es x = a · (y - k) ² + h

Con el vértice = (h, k)

Comparando, tenemos, h = -1 yk = 3, el vértice = (-1, 3)

También la ecuación de la parábola en forma cónica es (y - k) ² = 4 · p · (x - h)

Comparando con (y - 3) ² = -6 · (x + 1), tenemos 4p = -6, p = -3/2

El foco está en (h + p, k) que es (-1 + -3/2, 3) = (-5/2, 3)

Vértice = (-1, 3), Enfoque = (-5/2, 3)

3. Para la parábola, y = 12 · x, tenemos;

En comparación con la forma de la ecuación, y = a · x² + b · x + c

b = 12, a = 0, c = 0

Dado que el vértice = (h, k), tenemos;

h = -b / (2 × 0), h = ∞

k = a · h² + b · h + c = ∞

No hay vértice

Foco x valor = Vértice x valor = ∞

No hay foco

Directrix = (k - 1) / (4 · a) = (k - 1) / (4 × 0) = ∞, sin directriz

y = 12x no es una parábola

4. Para y = -8x, tampoco es una parábola como se muestra arriba.

4 0
3 years ago
How to write 13,180,000 in expanded form
Umnica [9.8K]
Thirteen million, one hundred eighty thousand.
5 0
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