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liubo4ka [24]
3 years ago
10

593,204 rounded to the nearest million

Mathematics
1 answer:
Romashka [77]3 years ago
6 0

The answer is  1,000,000

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Consider the function FX equals -2/3 X -24 which conclusions can be drawn about F negative 1X select two options a F negative 1X
omeli [17]

The conclusions that can be drawn about f–1(x) are:

  • f^-^1(x) has an x-intercept of (-36, 0)
  • f^-^1(x) has a range of all real numbers.

<h3>Calculations and Parameters:</h3>

Given:

f(x) = \frac{-2}{3}x - 24

f(x) + 24 = \frac{-2}{3}x

To find x, we would multiply both sides by -2/3

= x= \frac{-3}{2} (f(x) + 24)

f^-^1(x) = -\frac{3}{2} (0) - 36

f^-^1(x) = -36

Therefore, f^-^1(x)  has an x-intercept of (-36, 0)

Furthermore, f^-^1(x) has a range of all real numbers.

Read more about real numbers here:

brainly.com/question/17020343

#SPJ1

5 0
2 years ago
A number is chosen at random from 1 to 25. Find the probability of selecting an odd number or multiple of 5.
Sauron [17]
Odd number probability: 13/25 Multiple of five probability: 4/25
4 0
3 years ago
Read 2 more answers
Perform the indicated opera<br> 6) g(n)= n2 - 1<br> h(n)= 4n+1<br> Find (g. h)n)
DaniilM [7]

(g.h)(n)\,\,is\,\,4n^3+n^2-4n-1\\

Step-by-step explanation:

We are given:

g(n)=n^2-1\\h(n)=4n+1

We need to find (g.h)(n)

Finding (g.h)(n)

(g.h)(n)=g(n).h(n)\\ (g.h)(n)=(n^2-1)(4n+1)\\ (g.h)(n)=n^2(4n+1)-1(4n+1)\\ (g.h)(n)=4n^3+n^2-4n-1\\

So, (g.h)(n)\,\,is\,\,4n^3+n^2-4n-1\\

Keywords: Composite Function

Learn more about Composite Function at:

  • brainly.com/question/10772025
  • brainly.com/question/4939434
  • brainly.com/question/2723982

#learnwithBrainly

6 0
3 years ago
Pls answer correctly thank you
Marat540 [252]
Top left 127
top middle 57
top right 118

bottom left 30
bottom middle 65
bottom right 36
4 0
3 years ago
Answer the following true or false. Justify your answer.
Anna007 [38]

Answer:

a) False b) False c) True d) True e) True

Step-by-step explanation:

a) If A is a subset of B, and x∈B, then x∈A. False

Suppose A is Z (Set of Integers), and B is R (Set of Real Numbers). Then A is a subset of B. x ∈ B, let's say x equals π. If x ∈ B (B = Real Numbers) and x=π then x ∉ A (Z).

We could call A, any other subset of Real numbers (Q, I,..) and we both would come up to the same conclusion when it comes to Real numbers the Set.

So this conclusion is False. Not always an element of a subset is an element of a set.

b) False

For this one, I've drawn some lines, and it is useful to work with them.

Given the set {(x,y) ∈ R²| 0<x<0} then all negative and positive numbers but zero belong to this set.

c) If A and B are square matrices, then AB is also square. True

Taking into account the rules for multiplying matrices. The number of columns of A must be the same number of lines of B to there can be a matrix product.

Whenever we multiply square matrices, we'll always get square matrices. Then this conclusion is true.

d) A and B are subsets of a set S, then A∩B and A∪B are also subsets of S. True

Suppose A= Z (Integer Numbers) and B=Q (Rational Numbers) and S= R (Real Numbers)

A∩B = Z∩Q=∅ and A∪B =Z∪Q = subset

Since the ∅ empty set ⊂ in every set and ZUQ is another Subset of R this is a True conclusion.

e) True. For a matrix A, we define A²= AA

For any Power of Matrices, all we have to do is multiply any given matrix by itself for a given number of times.

M²=M*M

M³=M*M*M

7 0
3 years ago
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