Given f (x) = x2 + 3x – 4 and values of the linear function g(x) in the table, what is the range of (f + g)(x)?
x –6 –3 –1 4
g(x) 13 4 –2 –17
ℝ
[–3, 3]
(–∞, –9]
[–9, ∞)
Answers
The range of function (f + g)(x) is (-9, ∞) option (D) is correct the linear function g(x) is g(x) = -3x - 5
What is a function?It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.
We have a function:
f(x) = x² + 3x - 4
The domain of the f(x) is all real numbers
The range of the g(x) is [-6.25, ∞)
g(x) is shown in the table:
x –6 –3 –1 4
g(x) 13 4 –2 –17
The linear function g(x) is:
[tex]\rm g(x) = \ -13=\dfrac{\left(4-13\right)}{-3+6}\left(x+6\right)[/tex]
g(x) = -3x - 5
(f + g)(x) = f(x) + g(x)
= x² + 3x - 4 - 3x - 5
(f + g)(x) = x² - 9
The domain of the function is all real numbers.
The range of the function (-9, ∞)
Thus, the range of function (f + g)(x) is (-9, ∞) option (D) is correct the linear function g(x) is g(x) = -3x - 5
Learn more about the function here:
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Answers
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They packed 594 kilograms of rice.
Given:
Number of bags packed by the group of volunteers = 198
Number of kilo a bag contains = 3 kg
Therefore:
Number of kilograms of rice packed by the volunteers = number of bags packed [tex]\times[/tex] number of kilo one bag of rice weighs
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Number of kilos of rice packed = [tex]198 \times 3 $ kg\\[/tex]
[tex]= 594 $ kg[/tex]
Therefore, the volunteers packed 594 kilograms of rice.
Learn more here:
https://brainly.com/question/21146374
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Answers
Answers:
falsefalsetruefalseTrue==================================================
Explanation:
Problem 1
This is false because the A and B should swap places. It should be [tex](AB)^{-1} = B^{-1}A^{-1}[/tex].
The short proof is to multiply AB with its inverse [tex](AB)^{-1}[/tex] and we get: [tex](AB)*(AB)^{-1} = (AB)*(B^{-1}A^{-1}) = A(B*B^{-1})*A^{-1} = A*A^{-1} = I[/tex]
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-------------------------
Problem 2
This statement is true if and only if AB = BA
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-------------------------
Problem 3
This statement is true.
If A and B are invertible, then so is AB.
This is because both [tex]A^{-1}[/tex] and [tex]B^{-1}[/tex] are known to exist (otherwise A and B wouldn't be invertible) and we can use the rule mentioned in problem 1. Make sure to swap the terms of course.
Or you can use a determinant argument to prove the claim
det(A*B) = det(A)*det(B)
Since A and B are invertible, their determinants det(A) and det(B) are nonzero which makes the right hand side nonzero. Therefore det(A*B) is nonzero and AB has an inverse.
So if we have two invertible matrices, then their product is also invertible. This idea can be scaled up to include things like A^4*B^3 being also invertible.
If you wanted, you can carefully go through it like this:
If A and B are invertible, then so is ABIf A and AB are invertible, then so is A*AB = A^2BIf A and A^2B are invertible, then so is A*A^2B = A^3Band so on until you build up to A^4*B^3. Therefore, we can conclude that A^m*B^n is also invertible. Be careful about the order of multiplying the matrices. Something like A*AB is different from AB*A, the first of which is useful while the second is not.
So this is why statement 3 is true.
-------------------------
Problem 4
This is false. Possibly a quick counter-example is to consider these two matrices
[tex]A = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \text{ and } B = \begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix}[/tex]
both of which are invertible since their determinant is nonzero (recall the determinant of a diagonal matrix is simply the product along the diagonal entries). So it's not too hard to show that the determinant of each is 1, and each matrix shown is invertible.
However, adding those two mentioned matrices gets us the 2x2 zero matrix, which is a matrix of nothing but zeros. Clearly the zero matrix has determinant zero and is therefore not invertible.
There are some cases when A+B may be invertible, but it's not true in general.
-------------------------
Problem 5
This is true because each A pairs up with an [tex]A^{-1}[/tex] to cancel out (similar what happened with problem 1). For more info, check out the concept of diagonalization.