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44. the expression $23(5)^{x + 7}$ is equivalent to which of the follow…

Question

  1. the expression $23(5)^{x + 7}$ is equivalent to which of the following expressions?

a. $23^x(5)^7$
b. $23(5^x)^7$
c. $23(5)^x(5)^7$
d. $23(5^x) + 23(5)^7$

  1. a group of students are studying bacteria on their campus. one type of bacteria was found to grow in such a way that could be modeled by the equation $b(t) = 20(1.075)^{8t}$, where $b(t)$ is the number of bacteria after $t$ days.

which equation best represents an equivalent form of $b(t)$?
a. $b(t) = 20(1.78)^t$
b. $b(t) = 20(0.134)^t$
c. $b(t) = 160(1.78)^t$
d. $b(t) = 160(0.134)^t$

  1. choose an option for each blank to make the sentence true.

the choice 1 sequence 3, 12, 21, 30, 39, ... can be represented by the function choice 2, where $n$ is a positive whole number.

choice 1:

  • arithmetic
  • geometric

choice 2:

  • $f(n) = 4 + 3(n - 1)$
  • $f(n) = 3 + 4(n - 1)$
  • $f(n) = 9 + 3(n - 1)$
  • $f(n) = 3 + 9(n - 1)$
  1. in a physics class, liam learned that the electrical force between two charged bodies can be determined using the formula $f = \frac{kq_1q_2}{r^2}$, where $q_1$ and $q_2$ are the charges on the two objects, $r$ is the distance between the two objects and $k$ is the proportionality constant.

what is $r$ in terms of $f$, $k$, $q_1$ and $q_2$?
a. $r = \sqrt{\frac{q_1q_2}{kf}}$
b. $r = \sqrt{\frac{k}{fq_1q_2}}$
c. $r = \sqrt{\frac{k}{q_1q_2}}$
d. $r = \sqrt{\frac{kq_1q_2}{f}}$

Explanation:

Question 44

Step1: Usar la propiedad de exponentes $a^{m+n}=a^m \cdot a^n$

La expresión es $23(5)^{x + 7}$. Por la propiedad de exponentes, $5^{x+7}=5^x \cdot 5^7$.

Step2: Sustituir la propiedad en la expresión

Sustituyendo, $23(5)^{x + 7}=23 \cdot 5^x \cdot 5^7 = 23(5)^x(5)^7$.

Step1: Simplificar $(1.075)^{8t}$

Primero, calcular $1.075^8$. $1.075^8 \approx 1.78$ (usando una calculadora o propiedades de exponentes).

Step2: Reescribir la función $B(t)$

La función original es $B(t) = 20(1.075)^{8t}$. Sustituyendo $(1.075)^{8t}$ por $(1.78)^t$ (ya que $(a^m)^n=a^{mn}$ y $1.075^8 \approx 1.78$), tenemos $B(t)=20(1.78)^t$.

Step1: Determinar el tipo de secuencia (Choice 1)

Para la secuencia $3, 12, 21, 30, 39, \dots$, la diferencia entre términos consecutivos es $12 - 3 = 9$, $21 - 12 = 9$, $30 - 21 = 9$, $39 - 30 = 9$. Como la diferencia es constante, es una secuencia aritmética.

Step2: Encontrar la fórmula de la secuencia aritmética (Choice 2)

La fórmula para una secuencia aritmética es $f(n)=a_1 + d(n - 1)$, donde $a_1$ es el primer término y $d$ es la diferencia común. Aquí, $a_1 = 3$ y $d = 9$? No, espera, $d = 9$? Wait, $3, 12$: $12 - 3 = 9$? No, $12 - 3 = 9$? Wait, $3 + 9(n - 1)$: para $n = 1$, $3 + 9(0)=3$; $n = 2$, $3 + 9(1)=12$; $n = 3$, $3 + 9(2)=21$, lo que coincide. Wait, pero la opción es $f(n)=3 + 9(n - 1)$? Wait, las opciones de Choice 2: $f(n)=4 + 3(n - 1)$, $f(n)=3 + 4(n - 1)$, $f(n)=9 + 3(n - 1)$, $f(n)=3 + 9(n - 1)$. Wait, la diferencia es 9? Wait, $12 - 3 = 9$, $21 - 12 = 9$, sí. Entonces la secuencia es aritmética, y la fórmula es $f(n)=3 + 9(n - 1)$? Wait, no, $3 + 9(n - 1)$: para $n=1$, 3; $n=2$, 3 + 9(1)=12; $n=3$, 3 + 9(2)=21, correcto. Otra opción: $f(n)=3 + 4(n - 1)$: $n=2$ sería 3 + 4(1)=7, no coincide. $f(n)=4 + 3(n - 1)$: $n=1$ es 4, no coincide. $f(n)=9 + 3(n - 1)$: $n=1$ es 9, no coincide. Entonces Choice 1 es aritmética, Choice 2 es $f(n)=3 + 9(n - 1)$.

Answer:

C. $23(5)^x(5)^7$

Question 45