Ask question

Ask question

All categories

Svetllana [295]

3 years ago

13

What region was the location of a war between Russia and the allied nations of Britain in

1 answer:

blagie [28]3 years ago

7 0

Answer:

war fought mainly on the Crimean Peninsula between the Russians and the British, French, and Ottoman Turkish

Step-by-step explanation:

You might be interested in

Charlie saves $21.45 each month for 6 months.In the seventh month, he only saves $10.60.How much money will Charlie have after 7

Tomtit [17]

$21.45 x 6 = $128.7
$128.7 + $10.60 = $139.3

Charlie will have saved $139.3 in 7 months

8 0

3 years ago

A recent survey found that 3 out of every 7 people like birds. If 3500 people were surveyed, how many like birds?

photoshop1234 [79]

1500 hopefully that helps

8 0

3 years ago

Difference is 9, product is 22

expeople1 [14]

The answer is 2,444444444444444

3 0

3 years ago

Read 2 more answers

Is this equation true or false? 4(-5) = (-25)

aleksandr82 [10.1K]

4*-5 is -20 is does not equal -25 so it’s false

3 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