Answer:
Check the explanation
Step-by-step explanation:
kindly check the attached image below to Determine whether the given set S is a subspace of the vector space <u><em>(which is contained within a different vector space. So all the subspace is a kind of vector space in their own way, although it is also defined relative to some of the other larger vector space. The linear subspace is more often than not simply called a subspace whenever the situation serves to differentiate it from other types of subspaces.)</em></u> V.A
A) The domain remains the same.
Since, in y = - 2 f(x) , the y is whats being multiplied and therefore changed, not the x or domain.
Therefore D[-6,10] <span>
The range is what changes by a factor of -2.
If the original range was [-8,12], then the new one is
[(-8 x -2),(12 x -2)] = [16 , -24].
However, you must flip the numbers to make the interval true, giving you R [-24,16]</span>
Answer:
the answer is y=1/3x+20
Step-by-step explanation:
because he is going 3 times faster and you are adding on the 20 minutes that he is late
Answer:
P ′(−10, −2), L ′(−5, −4), K ′(−15, 11)
Step-by-step explanation:
The translation vector, ⟨−6,3⟩, can be used to derive a rule for the translation of the coordinates: (x,y)→(x−6,y+3).
Apply the rule to translate each of the three preimage vertices to the image vertices.
P(−4,−5)→P'(−10,−2)
L(1,−7)→L'(−5,−4)
K(−9,8)→K'(−15,11)
Therefore, P'(−10,−2), L'(−5,−4), N'(−15,11) is the translation of the figure with the vertices P(−4,−5), L(1,−7), and K(−9,8), along the vector ⟨−6,3⟩.
Answer:
D.) because it cannot be expressed as a ratio of integers
Step-by-step explanation:
The root of any integer that is not a perfect square is irrational. 5 is not a perfect square, so is irrational—it cannot be expressed as the ratio of integers.
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<em>Proof</em>
Suppose √5 = p/q, where p and q are mutually prime. Then p² = 5q².
If p is even, then q² must be even. We know that √2 is irrational, so the only way for q² to be even is for q to be even—contradicting our requirement on p and q.
If p is odd, then both p² and q² will be odd. We can say p = 2n+1, and q = 2m+1, so we have ...
p² = 5q²
(2n+1)² = 5(2m+1)²
4n² +4n +1 = 20m² +20m +5
4n² +4n = 4(4m² +4m +1)
n(n+1) = (2m+1)²
The expression on the left will be even for any integer n; the expression on the right will be odd for any integer m. This equation cannot be satisfied for any integers m and n, so contradicting our assumption √5 = p/q.
We have shown using "proof by contradiction" that √5 cannot be the ratio of integers.
