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3 years ago

I NEED SOMEONE TO HELP ME WITH MY TEST ITS ONLY 8 QUESTIONS. PLEASE PLEASE HELP HELP

Mathematics

2 answers:

photoshop1234 [79]3 years ago

3 0

Answer:

What kind of test????

Step-by-step explanation:

zimovet [89]3 years ago

3 0

What test ? type the questions i’ll see if i can help you

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F(x) =4x2 + 8x -5 factor completely

Burka [1]

The factors of 4x^2 +8x-5 are (2x-1) and (2x+5)

8 0

1 year ago

Simplify the rational expression. state any restrictions on the variable. n^4-11n^2+30/n^4-7n^2+10

Andrej [43]

For this case we have the following expression:
 $\frac{n ^ 4-11n ^ 2 + 30}{ n ^ 4-7n ^ 2 + 10} 

$
 We are going to make the following change of variables:
 $x = n ^ 2
$
Rewriting the expression we have:
 $\frac{(x ^ 2-11x + 30)}{(x ^ 2-7x + 10)} 

$
Factoring the expression we have:
 $\frac{(x-5) (x-6)}{(x-5) (x-6)} 
$
Canceling similar terms we have:
 $\frac{x-6}{x-2} 
$
Returning the change we have:
 $\frac{n ^ 2-6}{n ^ 2-2} 

$
 The resctrictions of the function are those that make the denominator equal to zero.
We have then:
 $n ^ 2-2 = 0

n = +/- \sqrt{2} 
$
 Answer:
The simplified expression is:
 $\frac{n ^ 2-6}{n ^ 2-2}$
The restrictions are:
 $n = \sqrt{2} 
$
$n = - \sqrt{2}$

7 0

2 years ago

¿que son los numeros naturales?

Inga [223]

Huh? Someone translate this for me plz

8 0

2 years ago

Read 2 more answers

A man is planning to take his vacation in three weeks. He determines that there is a 15% chance of rain on any given day in that

mixas84 [53]

Answer: 15%

Explanation: if it has 15% chance of rain on any given day so that’s 15% chance on every day that week and I’d think about rescheduling that vacation for another week

4 0

2 years ago

Determine the location and values of the absolute maximum and absolute minimum for given function : f(x)=(‐x+2)4,where 0<×&lt

brilliants [131]

Answer:

Where 0 < x < 3

The location of the local minimum, is (2, 0)

The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

$f'(x) = \dfrac{(-x + 2)^4}{dx} = -4 \cdot (-x + 2)^3 = 0$

∴ (-x + 2)³ = 0

x = 2

$f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx} = -12 \cdot (-x + 2)^2$

When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2

We have, f(2) = (-2 + 2)⁴ = 0

The location of the local minimum, is (2, 0)

Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

-x + 2 = 0, when x = 2

-x + 2 = -1, when x = 3

Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16

The location of the local maximum is at (0, 16).

5 0

2 years ago