All categories

3 years ago

Prove the inequality. c+ 1/c ≥2 for c>0

Mathematics

1 answer:

Verizon [17]3 years ago

8 0

Answer:

Step-by-step explanation:

Given c+1/c ≥ 2 for c> 0

Form a table for vales of c with corresponding values that are true to the equation

when c=1 the inequality will be

1+1/1 ≥ 2

2 ≥ 2 ----------------true

when c=2

2+1/2 ≥ 2

2.5 ≥ 2 -------------------true

when c=3

3+1/3 ≥ 2

3.33 ≥ 2

where c =-1

-1 + 1/-1 ≥ 2

-1 + -1 ≥ 2

-2 ≥ 2 --------------false

where c=-2

-2 + 1/-2 ≥ 2

-2.5 ≥ 2---------------false

Hence the inequality only holds for values of c>0

You might be interested in

Q(a+1) - 2q(a) if q(x) = x²+3x+4

siniylev [52]

$q(x) = {x}^{2} + 3x + 4$

_________________________________

Step(1)

To find q(a) we just need to put a instead of x in q(x) function.

Let's do it...

$q(a) = {a}^{2} + 3a + 4$

Multiply sides by -2 :

$- 2q(a) = - 2( {a}^{2} + 3a + 4)$

$- 2q(a) = - 2 {a}^{2} - 6a - 8$

_________________________________

Step (2)

To find q(a+1) we just need to put a+1 instead of x in q(x) function.

Let's do it...

$q(a + 1) = ({a + 1})^{2} + 3(a + 1) + 4 \\$

$q(a + 1) = {a}^{2} + 2a + 1 + 3a + 3 + 4 \\$

$q(a + 1) = {a}^{2} + 5a + 8$

_________________________________

Step (3)

$q(a + 1) - 2q(a) =$

${a}^{2} + 5a + 8 - 2 {a}^{2} - 6a - 8 = \\$

$- {a}^{2} - a = - a(a + 1)$

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

8 0

4 years ago

Can some one awser some of those please thank you please :)

Alexeev081 [22]

5. 1,000,000
6. 9,300
7. 5,070
8. 5, 280
9. 810
10. 220
11. 44,770
12. 76, 000

Those are a few answers for you. Try looking back at your notes (If you copied any) to figure out your problems :)

4 0

4 years ago

Read 2 more answers

Find the quotient -88, 11

irinina [24]

The answer is -88/11=-8

4 0

3 years ago

Read 2 more answers

What is the y-intercept of the function, and what does this tell you about the car?

scZoUnD [109]

<span>Write an equation in the standard form to represent the total rent (y) that Sam has to ... is the y-intercept of the function, and what does this tell you about the car?</span><span>
</span>

6 0

3 years ago

Navy PilotsThe US Navy requires that fighter pilots have heights between 62 inches and78 inches.(a) Find the percentage of women

Zigmanuir [339]

The first part of the question is missing and it says;

Use these parameters: Men's heights are normally distributed with mean 68.6 in. and standard deviation 2.8 in. Women's heights are normally distributed with mean 63.7 in. and standard deviation 2.9 in.

Answer:

A) Percentage of women meeting the height requirement = 72.24%

B) Percentage of men meeting the height requirement = 0.875%

C) Corresponding women's height =67.42 inches while corresponding men's height = 72.19 inches

Step-by-step explanation:

From the question,

For men;

Mean μ = 68.6 in

Standard deviation σ = 2.8 in

For women;

Mean μ = 63.7 in

Standard deviation σ = 2.9 in

Now let's calculate the standardized scores;

The formula is z = (x - μ)/σ

A) For women;

Z = (62 - 63.7)/2.9 = - 0.59

Z = (78 - 63.7)/2.9 = 4.93

The original question cam be framed as;

P(62 < X < 78).

So thus, the probability of only women will take the form of;

P(-0.59 < Z < 4.93) = P(Z<4.93) - P(Z > - 0.59)

From the normal probability table attached, when we interpolate, we'll arrive at P(Z<4.93) = 0.9999996

And P(Z > - 0.59) = 0.277595

Thus;

P(Z<4.93) - P(Z > - 0.59) =0.9999996 - 0.277595 = 0.7224

So, percentage of women meeting the height requirement is 72.24%.

B) For men;

Z = (62 - 68.6)/2.8 = -2.36

Z = (78 - 68.6)/2.8 = 3.36

Thus, the probability of only men will take the form of;

P(-2.36 < Z < 3.36) = P(Z<3.36) - P(Z > - 2.36)

From the normal probability table attached, when we interpolate, we'll arrive at P(Z<3.36) = 0.99961

And P(Z > -2.36) = 0.99086

Thus;

P(Z<3.36) - P(Z > -2.36) 0.99961 - 0.99086 = 0.00875

So, percentage of women meeting the height requirement is 72.24%.

B)For women;

Z = (62 - 63.7)/2.9 = - 0.59

Z = (78 - 63.7)/2.9 = 4.93

The original question cam be framed as;

P(62 < X < 78).

So thus, the probability of only women will take the form of;

P(-0.59 < Z < 4.93) = P(Z<4.93) - P(Z > - 0.59)

From the normal probability table attached, when we interpolate, we'll arrive at P(Z<4.93) = 0.9999996

And P(Z > - 0.59) = 0.277595

Thus;

P(Z<4.93) - P(Z > - 0.59) =0.9999996 - 0.277595 = 0.00875

So, percentage of women meeting the height requirement is 0.875%

C) Since the height requirements are changed to exclude the tallest 10% of men and the shortest10% of women.

For women;

Let's find the z-value with a right-tail of 10%. From the second table i attached ;

invNorm(0.90) = 1.2816

Thus, the corresponding women's height:: x = (1.2816 x 2.9) + 63.7= 67.42 inches

For men;

We have seen that,

invNorm(0.90) = 1.2816

Thus ;

Thus, the corresponding men's height:: x = (1.2816 x 2.8) + 68.6 = 72.19 inches

7 0

4 years ago