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const2013 [10]
3 years ago
13

00:00 Is the relation shown below a function? Use the graph and justify your answer (0, 0), (2.0), (2, 2), (3, 4) and (6,6) y 9

8 7 6 on 4 3 2 1 х 0 0 1 2 3 4 5 6 7 8 9 00:00 Yes, each x-value has a unique y-value 00:00​
Mathematics
1 answer:
Charra [1.4K]3 years ago
8 0

No; Two points have different y-values for the same x-value of 2.

Step-by-step explanation:

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If an employee works on a holiday, he or she is typically paid

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2 3/5 as a decimal.
GaryK [48]

Answer:

2 3/5 in decimal form will look like this.

2.6

7 0
3 years ago
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A baker makes 12 dozen
Zarrin [17]

Answer:

624

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8 0
3 years ago
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If h(x) = 5 x and k(x)=1/x, which expression is equivalent to (koh)(x)?
Lyrx [107]
May be there is an operator missing in the first function, h(x). I will solve this in two ways, 1) as if the h(x) = 5x and 2) as if h(x) = 5 + x

1) If h(x) = 5x and k(x) = 1/x

Then (k o h) (x) = k ( h(x) ) = k(5x) = 1/(5x)

2) If h(x) = 5 + x and k (x) = 1/x

Then (k o h)(x) =k ( h(x) ) = k (5+x) =  1 / [5 + x]
4 0
3 years ago
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(b) Find the angle between u and v to the nearest degree.
yulyashka [42]

<u>Question 1 (b)</u>

<u />u \cdot v=11\\\\|u|=\sqrt{3^{2}+(-1)^{2}}=\sqrt{10}\\\\|v|=\sqrt{4^{2}+1^{2}}=\sqrt{17}

So,

\cos \theta=\frac{11}{\sqrt{10}\sqrt{17}}\\\\\theta=\cos^{-1} \left(\frac{11}{\sqrt{10}\sqrt{17}} \right) \approx \boxed{32^{\circ}}

<u>Question 2 (b)</u>

<u />u \cdot v=22\\\\|u|=\sqrt{5^{2}+2^{2}}=\sqrt{29}\\\\|v|=\sqrt{4^{2}+1^{2}}=\sqrt{17}

So,

\cos \theta=\frac{22}{\sqrt{29}\sqrt{17}}\\\\\theta=\cos^{-1} \left(\frac{22}{\sqrt{29}\sqrt{17}} \right) \approx \boxed{8^{\circ}}

6 0
2 years ago
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