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2 years ago

How do i find the answer of this math question or problem f(x) = 4x − 9?

Mathematics

1 answer:

Mademuasel [1]2 years ago

6 0

Answer:

x = 9/4

Step-by-step explanation:

f(x) = 4x - 9

(first substitute f(x) with 0)

0 = 4× - 9

(Move four)

-4x = -9

=> x = -9/-4

= 9/4

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10c - cd + d^3 (if c = -6 and d= 2)

mash [69]

The required simplified solution of the given expression is -40.

Given that,
To determine the simplified solution of the given expression 10c - cd + d³ at c = -6 and d = 2.

<h3>What is simplification?</h3>

The process in mathematics to operate and interpret the function to make the function or expression simple or more understandable is called simplifying and the process is called simplification.

Here,
= 10c - cd + d³

Substitute the values,
= -60 + 12 + 8
= -40

Thus, the required simplified solution of the given expression is -40.

Learn more about simplification here:

brainly.com/question/12501526

#SPJ1

6 0

10 months ago

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

Six stand-up comics, A, B, C, D, E, and F, are to perform on a single evening at a comedy club. The order of performance is

kiruha [24]

Answer:

a) 1/6

b) 1/36

c) 1/720

d) 1/3

Step-by-step explanation:

a) Any of the six comics can perform in the fourth place, so there is one chance in six that Comic F is the one that performs fourth.

$P(F=4th)=1/6$

b) In this case, we have two conditions. Both are independent of each other, so the probability of both happening is the product of the probabilities of each happening individually:

$P(D=4th \&B=2nd)=P(D=4th)\cdot P(B=2nd)=(1/6)\cdot(1/6)=1/36$

c) This combination is one in all possible orders of perform. The amount of combinations of orders is n!=6!=720 possible combinations. So the probability of this specific order is:

$P(F, D, B, A, C, E)=1/720$

d) In this case, of the six possible comics performing in the last place, we calculate the probability of 2 of them being in that place. So the probability is:

$P(C=6th\,or\, E=6th)=2/6=1/3$