Ask question

Ask question

All categories

3 years ago

5

Last year Eric went on 6 business trips. The number of days of each was 5, 2, 8, 12, 6, and 3. How many weeks did Eric spend on

business trips last year

1 answer:

tatyana61 [14]3 years ago

7 0

I would say 5 weeks and 1 day but im not entirely sure. i just moved the numbers around. so 5 plus 2 equals 7 days which is one week. if you take 1 away from 8 it gives you 7 day, 2 weeks now. then i added that one day to the 6 to make 7 days, now 3 weeks. took away 5 from 12 to give me 7 days, 4 weeks now. then added that 5 to the 3 to make 8 days, 5 weeks and 1 day. i might have over complicated the problem but that is how i attacked it. hope it helps

You might be interested in

Please help with this question

Vladimir [108]

The answer is c because when you have 1 number and a variable(s) you just add the number to the variable(s)

6 0

3 years ago

Read 2 more answers

The sum of two number is 50,and their difference is 30.find the numbers

Mrac [35]

Answer:

40 and 10

Step-by-step explanation:

if you add them you get 50 and if you subtract them you get 30

6 0

3 years ago

Read 2 more answers

Due to a membership drive for a public television station, the current membership is 125% of what it was a year ago. The current

Oliga [24]

$x(1.25)=1200$

$\frac{x(1.25)}{1.25}=\frac{1200}{1.25}$

$x=960$

8 0

3 years ago

Read 2 more answers

Aldon and Jamal raised the same amount of money for the school fundraiser. Aldon

Naddika [18.5K]

Answer: $5

Step-by-step explanation: This can be represented by the equation

40 + 12T = 25 + 15T

We just solve for T.

Subtract 12T from both sides

40 = 25 + 3T Subtract 25

15 = 3T Divide by 3

5 = T

7 0

3 years ago

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

8 0

3 years ago