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My name is Ann [436]
3 years ago
11

Rewrite the following problems using exponent rules. Then, solve using logarithms.

Mathematics
1 answer:
Anna35 [415]3 years ago
3 0

9514 1404 393

Answer:

  1. 3(0.96059601^t) = 15; t = -40.03

  2. 4.121204(1.01^t) = 8; t = 66.66

Step-by-step explanation:

The applicable rules of exponents are ...

  (a^b)^c = a^(bc)

  (a^b)(a^c) = a^(b+c)

The solution for an exponential problem using logarithms is ...

  a·b^x = c   ⇒   log(c/a)/log(b) = x

_____

1. Rewrite: (3)(0.99^4)^t = 15  ⇒  3(0.96059601^t) = 15

Solution: t = log(15/3)/log(0.96059601) ≈ -40.03

__

2. Rewrite: 4(1.01^3)(1.01^t) = 8  ⇒  4.121204(1.01^t) = 8

Solution: t = log(8/4.121204)/log(1.01) ≈ 66.66

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o.42

Step-by-step explanation:

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4 years ago
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Simplify √3×√48is the answer
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8 0
3 years ago
Marisol noticed that she does not have a common factor.
konstantin123 [22]

Answer:  option 2 describes best

Step-by-step explanation:Given Marisol grouped the terms and factored the GCF out of the groups of the polynomial 6x3 – 22x2 – 9x + 33. Her work is shown.  

Step 1: (6x3 – 22x2) – (9x + 33)

Step 2: 2x2(3x – 11) – 3(3x + 11)

Marisol noticed that she does not have a common factor. Which accurately describes what Marisol should do next?

Marisol should realize that her work shows that the polynomial is prime.

Marisol should go back and group the terms in Step 1 as (6x3 – 22x2) – (9x – 33).

Marisol should go back and group the terms in Step 1 as (6x3 – 22x2) + (9x – 33).

Marisol should refactor the expression in Step 2 as 2x2(3x + 11) – 3(3x + 11).

According to question Marisol grouped the terms and has done factorisation of the given polynomial 6x^3 – 22x^2 – 9x + 33.

In step 1 she has written as (6x^3 – 22x^2) – (9x + 33)

Marisol has to go to step 1 in order to correct her mistake. She has to group the expression as (6x^3 – 22x^2) – (9x – 33) so that she will be able to get the expression as

6x^3 – 22x^2 – 9x + 33 after opening the brackets.

4 0
3 years ago
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Common Core - How many thousands are in. 273,050? explain your answer
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7 0
4 years ago
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For what value of a does (1/7)^3a+3=343^a-1?
CaHeK987 [17]

Answer: ok, remember that when

x^a=x^b, when x=x, then a=b

also

x^-m=1/(x^m)

and

(x^m)^n=x^(mn)

so

we can rewrite 1/7 as 7^-1

so

and 343=7^3

7=7 so

-3a-3=3a-3

add 3a to both sides

-3=6a-3

add 3 to both sides

0=6a

0=a

a=0

Answer: 0

By: apologiabiology

5 0
4 years ago
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