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3 years ago

Answer two questions about Equations A and B:

Mathematics

2 answers:

Gnesinka [82]3 years ago

8 0

Answer:

Jake interphalangeal the same time I was waiting to hear that your company and they will have the same

Scilla [17]3 years ago

8 0

Answer: A

2x-1=5x

-2x -2x

-1=3x

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Which value cannot represent the probability of an event occurring?

andrew-mc [135]

Answer:

1.1

Step-by-step explanation:

1 in probability means always so anything greater than 1 is meaningless

6 0

3 years ago

Consider the following theorem and proof.Theorem: The number âš2 is not rational number.Proof: Let's suppose âš2 is a rational n

ololo11 [35]

Answer:

The statement "Both of the numbers a and b cannot be even." is justified by the fact that a/b is simplified lowest terms

Step-by-step explanation:

We need to show that the √2 is an irrational number.

And from the given steps of proof stated in the question, we need to find the assumption that justifies the fact : " Both of the number a and b cannot be even".

First take the given options :

Option a : √2 is a rational number

√2 being an rational or irrational has no relation of a and b to be even or odd. So, this option is rejected.

Option B : a/b is simplified lowest terms

This shows that a and b are not even because if a and b are even then a/b can be simplified in other lowest term.

Option c : √2 is a irrational number

Similarly, By using the inverse part of Option A, option c is also rejected.

Option d : The fact that b divides a evenly

This only shows that the a is even. This does not give any idea about b is even or not. So option D is also rejected.

Option E : The fact that a and b are whole numbers

This fact does not imply that the a and b are even or odd. So option E is also rejected.

Hence, The statement "Both of the numbers a and b cannot be even." is justified by the fact that a/b is simplified lowest terms

7 0

2 years ago

WILL MAKE BRAINLIEST!!

kkurt [141]

I think it is -2, if that helps

4 0

3 years ago

What is 16million ×64 no using calculator

SIZIF [17.4K]

one billion twenty-four million

6 0

3 years ago

Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to f

tino4ka555 [31]

Answer:

The minimum value of the given function is f(0) = 0

Step-by-step explanation:

Explanation:-

Extreme value :- f(a, b) is said to be an extreme value of given function 'f' , if it is a maximum or minimum value.

i) the necessary and sufficient condition for f(x) to have a maximum or minimum at given point.

ii) find first derivative $f^{l} (x)$ and equating zero

iii) solve and find 'x' values

iv) Find second derivative $f^{ll}(x) >0$ then find the minimum value at x=a

v) Find second derivative $f^{ll}(x)$ then find the maximum value at x=a

Problem:-

Given function is f(x) = log ( x^2 +1)

step1:- find first derivative $f^{l} (x)$ and equating zero

$f^{l}(x) = \frac{1}{x^2+1} \frac{d}{dx}(x^2+1)$

$f^{l}(x) = \frac{1}{x^2+1} (2x)$ ……………(1)

$f^{l}(x) = \frac{1}{x^2+1} (2x)=0$

the point is x=0

step2:-

Again differentiating with respective to 'x', we get

$f^{ll}(x)=\frac{x^2+1(2)-2x(2x)}{(x^2+1)^2}$

on simplification , we get

$f^{ll}(x) = \frac{-2x^2+2}{(x^2+1)^2}$

put x= 0 we get $f^{ll}(0) = \frac{2}{(1)^2}$ > 0

$f^{ll}(x) >0$ then find the minimum value at x=0

Final answer:-

The minimum value of the given function is f(0) = 0

5 0

3 years ago