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3 years ago

The excluded values of a rational expression are 2 and 5. Which of the following could be this expression?

Mathematics

1 answer:

LuckyWell [14K]3 years ago

3 0

Answer:

You have not added the options, therefore, I cannot provide an exact answer. However, I can help you with the concept.

In any fraction, the denominator cannot be equal to zero because this would make the fraction undefined.

Excluded values are the values that would make the denominator equal to zero.

We are given that the excluded values are 2 and 5, this means that the factors of the denominator are (x-2) and (x-5)

This means that the denominator is x² - 7x + 10

Pick the fraction which has this denominator.

Hope this helps :)

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A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat

jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.

We are given a function (corrected)

$N(t) = 20(t^2-lnt^2)+ 30$

$N(t) = 20(t^2-2lnt)+ 30$

(a)

First, we take its derivative

$N'(t) = 20(2t-\frac{2}{t})$

Solve N'(t)=0

$20(2t-\frac{2}{t})=0$

Simplifying

$2t^2-2=0$

Solving for t

$t=1\ ,t=-1$

Only t=1 belongs to the valid interval $1\leqslant t\leqslant 15$

Taking the second derivative

$N''(t) = 20(2+\frac{2}{t^2})$

Which is always positive, so t=1 is a minimum

(b)

$N(1)=20(1^2-2ln1)+ 30$

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

$N(15)=20(15^2-2ln15)+ 30$

N(15)=4391.7 is a maximum

7 0

3 years ago

The indoor climbing gym is a popular field trip destination. This year the senior class at High School A and the senior class at

Tamiku [17]

Answer:

50 students per bus and 9 students per van

Step-by-step explanation:

I just guessed until I got it right, so I don't really have a method. If I ever find a better way of doing it, I'll edit my answer.

7 0

3 years ago

Help please!!!!!!!!!

Alenkasestr [34]

Answer:

B. $x^{2}$ - 4x - 5

Step-by-step explanation:

(p о n) (x) is the same thing as (p(n(x))

We know n(x) = x - 5, so let's input that into our expression

(p(x - 5))

Since we know p(x) = x^2 + 6x, let's replace the x with the new value x-5. Now our expression is:

p(x) = (x - 5)^2 + 6(x - 5)

Now all we have to do is simplify.

(x - 5)^2 is the same as: (x - 5)(x - 5)

Using the foil method, it simplifies to: x^2 - 10x + 25

6 (x - 5) = 6x - 30

Now our expression is: x^2 - 10x + 25 + 6x - 30

Combine like terms: x^2 - 4x - 5

The expression is: $x^{2}$ - 4x - 5

5 0

2 years ago

Plz help brainiest to whoever right

nirvana33 [79]

Answer:

X=11

m<wxz= 23

Step-by-step explanation:

hope this helps branliest please

3 0

2 years ago

Christian sets aside $60 per month for daily coffee. McDonalds sells coffee for $1, and The Bean Sprout sells a cup of coffee fo

VladimirAG [237]

Since Christian sets aside $60 per month for daily coffee, and McDonalds sells coffee for $ 1, and The Bean Sprout sells a cup of coffee for $3, to create an equation that represents the combinations of coffee options that allow Christian to stay in his budget and provide 2 combinations that use his entire budget, 1 combination that would be under budget, and 1 combination that would be over his budget, the following linear equation must be proposed:

M + BS = 60

2 combinations using the entire budget =

15 McDonalds + 15 The Bean Sprout = 15 + 45 = 60
30 McDonalds + 10 The Bean Sprout = 30 + 30 = 60

1 combination under budget =

15 McDonalds + 10 The Bean Sprout = 15 + 30 = 45

1 combination over budget =

30 McDonalds + 30 The Bean Sprout = 30 + 90 = 120

Learn more in brainly.com/question/14305871

4 0

2 years ago