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iren2701 [21]
3 years ago
6

Find the X-intercept f(x)=3^(x-1)-2

Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
7 0
<h2>Hello!</h2>

The answer is:

The function intercepts de x-axis at "x" equal to 1.631.

<h2>Why?</h2>

To find the x-intercept of the function, we need to make f(x) or "y" equal to 0, and then, isolate "x".

So, isolating we have:

f(x)=3^{(x-1)} -2\\\\0=3^{(x-1)} -2\\\\3^{(x-1)}=2\\\\Log_{3}(3^{(x-1)})=Log_{3}(2)\\\\x-1=0.631\\\\x=0.631+1=1.631

Hence, we have that the function intercepts de x-axis at "x" equal to 1.631.

Have a nice day!

Note: I have attached a picture for better understanding.

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Answer:

35

Step-by-step explanation:

by adding three to the next number each time, the answer is 35

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kakasveta [241]

Answer:

y = -1/4x + 3

Step-by-step explanation:

Because you are finding the perpendicular slope, you need to find the negative reciprocal of the original line, which would be -1/4. You then use point slope form to find the y-intercept with the slope and given point:

y - 5 = -1/4(x + 8). That equals to y = -1/4x + 3.

So the equation of this line is y = -1/4x + 3.

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2 years ago
Which equation does not have the same solution as the others?
ladessa [460]
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x - 9 = 17....x = 26 <=== there is the odd ball
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4 0
3 years ago
If the following fraction is reduced, what will be the exponent on the p? -5p^5q^4/ 8p^2q^2
viktelen [127]
<span>-5p^5q^4/ 8p^2q^2
= -5p^3q^2

</span><span>exponent on the p will be 3

answer
</span><span>C.3</span>
6 0
3 years ago
Read 2 more answers
La barbería El Caleño, tiene en promedio 120 clientes a la semana a
Luba_88 [7]

Queremos maximizar el precio de tal forma que los ingresos no disminuyan.

Ese maximo precio es: $14,040.6

Sabemos que actualmente el precio es:

p = $6,000

El número de clientes es:

C = 120

Actualmente los ingresos son el producto de esos dos números, es decir:

ingresos = $6,000*120 = $720,000

Ahora sabemos que por cada incremento de $700 en el precio, el número de clientes decrece en 10.

Entonces podemos escribir el número de clientes como una ecuación lineal.

C(p) = a*p + b

tal que tenemos dos puntos en esa linea:

($6,000, 120)

($6,700, 110)

La pendiente es:

a = \frac{110 - 120}{\$6,700 - \$6,000} = \frac{-10}{\$ 700}

Entonces tenemos:

C(p) = (-10/$700)*p + b

Sabemos que:

C($6,000) = 120 = (-10/$700)*$6,000 + b

                     120 = -85.71 + b

                     120 + 85.71 = b =

Entonces la ecuación lineal es:

C(p) = (-10/$700)*p + 205.71

Los ingresos serán dados por:

ingresos = C(p)*p = (-10/$700)*p^2 + 205.71*p

Y queremos maximizar p de tal forma que esto sea igual a lo que obtuvimos antes:

(-10/$700)*p^2 + 205.71*p = $720,000

Entonces debemos resolver la ecuación cuadratica:

(-10/$700)*p^2 + 205.71*p - $720,000 = 0.

Las soluciones son dadas por la formula de Bhaskara.

p = \frac{-205.71 \pm \sqrt{(205.71)^2 - 4*(-10/\$ 700)*\$ 720,000} }{2*(-10/\$ 700)} \\\\p = \frac{-205.71 \pm 195.45}{(-20/\$ 700)}

La solución de maximo valor es:

p = (-205.71 - 195.45)/(-20/$700) = $14,040.6

Sí quieres aprender más, puedes leer.

brainly.com/question/8926135

7 0
2 years ago
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