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3 years ago

What is the value of the expression below when n = 2.4 ? 4n X 2 + 43

Mathematics

1 answer:

meriva3 years ago

3 0

4(2.4)(2)+43
(9.6)(2)+43
19.2+43
62.2
the value of 4n x 2 + 43, when n=2.4, is 62.2

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Find the extreme values of the function and where they occur. f(x)=x-4sqrt(x)

Tema [17]

The extreme values of the function f(x) = x - 4sqrt(x) occur at the values of x for which f'(x) = 0

f'(x) = 1 - 2/sqrt(x) = 0
sqrt(x) - 2 = 0
sqrt(x) = 2
x = 4
To check for minimum or maximum,
f''(x) = 1/(sqrt(x))^3 = 1/(sqrt(4))^3 = 1/(2^3) = 1/8 => the point is a minimum.

Therefore, the minimum value = f(4) = 4 - 4sqrt(4) = 4 - 4(2) = 4 - 8 = -4 and occurs at x = 4.

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3 years ago

Number of

jekas [21]

Answer:

The constant rate of change is 3

Step-by-step explanation:

Given

x y

2 6

4 12

6 18

8 24

Required

Determine the constant rate of change

Represent the constant rate of change with k.

k is calculated using:

$k = \frac{y}{x}$

When y = 24; x = 8

So, we have:

$k = \frac{24}{8}$

$k = 3$

7 0

3 years ago

Find the sum of the geometric series.<br> 8+6+9/2+27/8+...

Tema [17]

Answer: The answer is 58. Hope this helps!

3 0

3 years ago

Read 2 more answers

Verify cot x sec^4x=cotx +2tanx +tan^3x

Tanzania [10]

Answer:

See explanation

Step-by-step explanation:

We want to verify that:

$\cot(x) \: { \sec}^{4} x = \cot(x) + 2 \tan(x) + { \tan}^{3} x$

Verifying from left, we have

$\cot(x) \: { \sec}^{4} x = \cot(x) \: ( 1 + { \tan}^{2} x )^{2}$

Expand the perfect square in the right:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: ( 1 + { 2\tan}^{2} x + { \tan}^{4} x)$

We expand to get:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: + \cot(x){ 2\tan}^{2} x +\cot(x) { \tan}^{4} x$

We simplify to get:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: + 2 \frac{ \cos(x) }{\sin(x) ) } \times \frac{{ \sin}^{2} x}{{ \cos}^{2} x} +\frac{ \cos(x) }{\sin(x) ) } \times \frac{{ \sin}^{4} x}{{ \cos}^{4} x}$

Cancel common factors:

$\cot(x) \: { \sec}^{4} x = \cot(x) \: + 2 \frac{{ \sin}x}{{ \cos}x} +\frac{{ \sin}^{3} x}{{ \cos}^{3} x}$

This finally gives:

$\cot(x) \: { \sec}^{4} x = \cot(x) + 2 \tan(x) + { \tan}^{3} x$

3 0

4 years ago

I WILL MARK FIRST BRAINLIEST! I leave a thanks too! 0 = -2x^2 - 7x -1 Solve using quadratic formula PLEASEEEEEE I need the answe

slamgirl [31]

Here,

2x^2 + 7x + 1 = 0

given,

a = 2

b = 7

and c = 1

From formula,

x = (-b+-√b^2-4ac) / 2a.

We get,

x = -7/4 ± √(41)/4 Ans.

4 0

3 years ago