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AveGali [126]
3 years ago
11

Any suggestions or answers

Mathematics
2 answers:
lyudmila [28]3 years ago
8 0
I think the answer is B
ryzh [129]3 years ago
6 0

I believe the answer is B.) 9and27/32

Best of luck.

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Local extreme Value for each of the Following A) F(x)=x^² - 4x² +5​
kobusy [5.1K]

Answer:

max{x²-4x²+5} = 5 at x = 0

Step-by-step explanation:

1. Find the critical numbers by finding the first derivative of f(x), set it to 0 and solve for x.

f'(x)=0

We get:

f(x) = -3x^2+5\\f'(x) = -6x\\-6x = 0\\x = 0

So the critical number is x = 0.

2. Evaluate the first derivative by plugging in the critical number and see if the derivative is positive or negative on both sides:

f'(x) is positive when the x < 0 (for example: -6*(-1)=+)

f'(x) is negative when the x > 0 (for example: -6*(1)=-)

Therefore, you have a local maximum.

Now just get the Y value by plugging in the critical number in the original function. f(0)=5

local maximum is (0,5)

4 0
1 year ago
Larry fills his bathtub at a constant rate. The amount of water in his tub is proportional to the amount of time he spends filli
Kryger [21]

Answer:

7 but im not sure

Step-by-step explanation:

8 0
3 years ago
Una caja de 25 newtons se suspende mediante un cable con diámetro de 2cm¿cuál es el esfuerzo aplicado al cable?
Naddik [55]

El cable experimenta un esfuerzo axial de 79577.472 pascales por el peso de la caja.

<h3>¿Cómo calcular el esfuerzo aplicado sobre el cable?</h3>

La caja tiene masa y está sometida a un campo gravitacional, por tanto, tiene un peso (W), en newtons. Por el principio de acción y reacción (tercera ley de Newton), encontramos que el cable es tensionado debido a ese peso y su área transversal experimenta un esfuerzo axial (σ), en pascales.

Asumiendo una distribución uniforme de la fuerza sobre toda la superficie transversal de la cuerda, tenemos que el esfuerzo axial se calcula mediante la siguiente expresión:

σ = W / (π · D² / 4)

Donde:

  • W - Peso de la caja, en newtons.
  • D - Diámetro del área transversal de la caja, en metros.

Si sabemos que W = 25 N y D = 0.02 m, entonces el esfuerzo axial aplicado a la cuerda es:

σ = 25 N / [π · (0.02 m)² / 4]

σ ≈ 79577.472 Pa

<h3>Observación</h3>

La falta de problemas verificados en español sobre esfuerzos axiales obliga a buscar uno equivalente en inglés.

Para aprender más sobre esfuerzos axiales: brainly.com/question/13683145

#SPJ1

5 0
1 year ago
9 is 9% of what number?
hjlf

Answer:

are you sure this is high school level stuff? its 9% of 100

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Company A: $21 per jersey<br>company B: $18 per jersey+$45 fee. when is company a best?
Anni [7]
So I would first find out when they are equal then do one more than that. 21x=18x+45, 3x=45, x=15 so the answer would be 16 jerseys
7 0
3 years ago
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