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3 years ago

ANSWER ASAP PLEASE AND SHOW YOUR WORK!!!!

Mathematics

1 answer:

IgorLugansk [536]3 years ago

5 0

Answer:

f(-x)= -f(x) so, the given function $f(x)= x^3-2x$ is Odd

Step-by-step explanation:

We need to determine if the function $f(x)= x^3-2x$ is Even, Odd, or Neither.

Determining if a function f(x) is even or not we put x=-x and check

If f(-x)=f(x) the function is even

if f(-x)= -f(x) the function is odd

So, Putting x =-x and determining if the given function $f(x)= x^3-2x$ is Even, Odd, or Neither.

$f(x)= x^3-2x\\Put \ x=-x\\f(-x)=(-x)^3-2(-x)\\f(-x)=-x^3+2x\\f(-x)=-(x^3-2x)\\f(-x)=-f(x)$

As f(-x)= -f(x) so, the given function $f(x)= x^3-2x$ is Odd

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Show solutions please

slega [8]

Answer:

1. Their ages are:

Steve's age = 18

Anne's age = 8

2. Their ages are:

Max's age = 17

Bert's age = 11

3. Their ages are:

Sury's age = 19

Billy's age = 9

4. Their ages are:

The man's age = 30

His son's age = 10

Step-by-step explanation:

1. We make the assumption that:

S = Steve's age

A = Anne's age

In four years, we are going to have:

S + 4 = (A + 4)2 - 2 = 2A + 8 - 2

S + 4 = 2A + 6 .................. (1)

Three years ago, we had:

S - 3 = (A - 3)3

S - 3 = 3A - 9

S = 3A - 9 + 3

S = 3A – 6 …………. (2)

Substitute S from (2) into (1) and solve for A, we have:

3A – 6 + 4 = 2A + 6

3A – 2A = 6 + 6 – 4

A = 8

Substitute A = 8 into (3), we have:

S = (3 * 8) – 6 = 24 – 6

S = 18

Therefore, we have:

Steve's age = 18

Anne's age = 8.

2. We make the assumption that:

M = Max's age

B = Bert's age

Five years ago, we had:

M - 5 = (B - 5)2

M - 5 = 2B - 10 .......................... (3)

A year from now, it will be:

(M + 1) + (B + 1) = 30

M + 1 + B + 1 = 30

M + B + 2 = 30

M = 30 – 2 – B

M = 28 – B …………………… (4)

Substitute M from (4) into (3) and solve for B, we have:

28 – B – 5 = 2B – 10

28 – 5 + 10 = 2B + B

33 = 3B

B = 33 / 3

B = 11

If we substitute B = 11 into equation (4), we will have:

M = 28 – 11

M = 17

Therefore, their ages are:

Max's age = 17

Bert's age = 11.

3. We make the assumption that:

S = Sury's age

B = Billy's age

Now, we have:

S = B + 10 ................................ (5)

Next year, it will be:

S + 1 = (B + 1)2

S + 1 = 2B + 2 .......................... (6)

Substituting S from equation (5) into equation (6) and solve for B, we will have:

B + 10 + 1 = 2B + 2

10 + 1 – 2 = 2B – B

B = 9

Substituting B = 9 into equation (5), we have:

S = 9 + 10

S = 19

Therefore, their ages are:

Sury's age = 19

Billy's age = 9.

4. We make the assumption that:

M = The man's age

S = His son's age

Therefore, now, we have:

M = 3S ................................... (7)

Five years ago, we had:

M - 5 = (S - 5)5

M - 5 = 5S - 25 ................ (8)

Substituting M = 3S from (7) into (8) and solve for S, we have:

3S - 5 = 5S – 25

3S – 5S = - 25 + 5

-2S = - 20

S = -20 / -2

S = 10

Substituting S = 10 into equation (7), we have:

M = 3 * 10 = 30

Therefore, their ages are:

The man's age = 30

His son's age = 10

3 0

3 years ago

What is the value of (-14^0)^-2?

Ainat [17]

Answer:

-6

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Simplify. (11 + (+4 + 1)2] - 11 1-1 - 21

Sergeu [11.5K]

The chosen topic is not meant for use with this type of problem.

6 0

3 years ago

If you are given a3=2 a5=16, find a100.

ra1l [238]

I suppose $a_n$ denotes the n-th term of some sequence, and we're given the 3rd and 5th terms $a_3=2$ and $a_5=16$ . On this information alone, it's impossible to determine the 100th term $a_{100}$ because there are infinitely many sequences where 2 and 16 are the 3rd and 5th terms.

To get around that, I'll offer two plausible solutions based on different assumptions. So bear in mind that this is not a complete answer, and indeed may not even be applicable.

• Assumption 1: the sequence is arithmetic (a.k.a. linear)

In this case, consecutive terms differ by a constant d, or

$a_n = a_{n-1} + d$

By this relation,

$a_{n-1} = a_{n-2} + d$

and by substitution,

$a_n = (a_{n-2} + d) + d = a_{n-2} + 2d$

We can continue in this fashion to get

$a_n = a_{n-3} + 3d$

$a_n = a_{n-4} + 4d$

and so on, down to writing the n-th term in terms of the first as

$a_n = a_1 + (n-1)d$

Now, with the given known values, we have

$a_3 = a_1 + 2d = 2$

$a_5 = a_1 + 4d = 16$

Eliminate $a_1$ to solve for d :

$(a_1 + 4d) - (a_1 + 2d) = 16 - 2 \implies 2d = 14 \implies d = 7$

Find the first term $a_1$ :

$a_1 + 2\times7 = 2 \implies a_1 = 2 - 14 = -12$

Then the 100th term in the sequence is

$a_{100} = a_1 + 99d = -12 + 99\times7 = \boxed{681}$

• Assumption 2: the sequence is geometric

In this case, the ratio of consecutive terms is a constant r such that

$a_n = r a_{n-1}$

We can solve for $a_n$ in terms of $a_1$ like we did in the arithmetic case.

$a_{n-1} = ra_{n-2} \implies a_n = r\left(ra_{n-2}\right) = r^2 a_{n-2}$

and so on down to

$a_n = r^{n-1} a_1$

Now,

$a_3 = r^2 a_1 = 2$

$a_5 = r^4 a_1 = 16$

Eliminate $a_1$ and solve for r by dividing

$\dfrac{a_5}{a_3} = \dfrac{r^4a_1}{r^2a_1} = \dfrac{16}2 \implies r^2 = 8 \implies r = 2\sqrt2$

Solve for $a_1$ :

$r^2 a_1 = 8a_1 = 2 \implies a_1 = \dfrac14$

Then the 100th term is

$a_{100} = \dfrac{(2\sqrt2)^{99}}4 = \boxed{\dfrac{\sqrt{8^{99}}}4}$

The arithmetic case seems more likely since the final answer is a simple integer, but that's just my opinion...

3 0

2 years ago

HELPPP !! If O N = 8 x − 8 , L M = 7 x + 4 , N M = x − 5 , and O L = 3 y − 6 , find the values of x and y for which LMNO must be

sveticcg [70]

Answer:

x=12 and y=4.3

Step-by-step explanation:

Given the sides of quadrilateral that are O N = 8 x − 8 , L M = 7 x + 4 , N M = x − 5 , and O L = 3 y − 6. we have to find the value of x and y so that LMNO be a parallelogram.

we know opposite sides of parallelogram are equal.

Hence, we have to find the value of x and y such that opposite sides becomes equal which implies LMNO is a parallelogram.

Equating opposite sides equal, we get

8x-8=7x+4 ⇒ 8x-7x=8+4=12 ⇒ x=12

implies, NM=x-5=12-5=7

NM=OL ⇒ 3y-6=7 ⇒ 3y=13 ⇒ y=4.3

Hence, x=12 and y=4.3

4 0

3 years ago