Which statement is correct about a lime and a point?

Find all solutions

Anton [14]

Answer:

The solutions of the equation are 0 , π

Step-by-step explanation:

* Lets revise some trigonometric identities

- sin² Ф + cos² Ф = 1

- tan² Ф + 1 = sec² Ф

* Lets solve the equation

∵ tan² x sec² x + 2 sec² x - tan² x = 2

- Replace sec² x by tan² x + 1 in the equation

∴ tan² x (tan² x + 1) + 2(tan² x + 1) - tan² x = 2

∴ tan^4 x + tan² x + 2 tan² x + 2 - tan² x = 2 ⇒ add the like terms

∴ tan^4 x + 2 tan² x + 2 = 2 ⇒ subtract 2 from both sides

∴ tan^4 x + 2 tan² x = 0

- Factorize the binomial by taking tan² x as a common factor

∴ tan² x (tan² x + 2) = 0

∴ tan² x = 0

OR

∴ tan² x + 2 = 0

∵ 0 ≤ x < 2π

∵ tan² x = 0 ⇒ take √ for both sides

∴ tan x = 0

∵ tan 0 = 0 , tan π = 0

∴ x = 0

∴ x = π

OR

∵ tan² x + 2 = 0 ⇒ subtract 2 from both sides

∴ tan² x = -2 ⇒ no square root for negative value

∴ tan² x = -2 is refused

∴ The solutions of the equation are 0 , π

4 0

3 years ago

Read 2 more answers

In 2010, the population of a city was 241,000. From 2010 to 2015, the population grew by 5%. From 2015 to 2020, it fell by 4%. T

lawyer [7]

Answer:

242, 900 people

Step-by-step explanation:

To begin, it says that the population grew by 5%.

(241000)(1.00 + 0.05) = population of the city in 2015

(241000)(1.05) = 253050

253050 is the population of the city in 2015. From 2015 to 2020 it fell four percent.

(253050)(1.00 - 0.04) = population of the city in 2020

(253050)(0.96) = 242928

242928 is the exact population of the city. However, it says to round to the nearest hundred. Since 28 is less than 50, then it would round down to 900.

Therefore, the answer would be 242, 900 people.

4 0

2 years ago

How do i do this? And whats the answer

sveticcg [70]

Step-by-step explanation:

just guess and see

7 0

2 years ago

Help i will give brainlist

Arisa [49]

Answer:

3. 2/10 4/20 3/15

4. 12/1 48/4 36/3

Step-by-step explanation:

3 0

2 years ago

Prove the following by induction. In each case, n is apositive integer. 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

To prove: $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago