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trasher [3.6K]
3 years ago
10

What is the domain of the function continous or discrete

Mathematics
1 answer:
scoray [572]3 years ago
6 0
The domain of the function is discrete 

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True or False?
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#1

That is false..one could have side lengths of 9 and 1, and the other could have side lengths of 3. Both the areas would be 9, but the figures would not be congruent.

#2

That is true, they must both have the same side length to have the same perimeter, therefore they will also have the same area.
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3 years ago
Solve the proportion for x.
Delvig [45]

\frac{x}{8}  =  \frac{8}{64}  \\ 64x = 8 \times 8 \\ 64x = 64 \\ x =  \frac{64}{64}  \\  \boxed{x = 1}

  • Proportion for x is <u>1</u><u>.</u>
  • Use <u>cross </u><u>multiplication</u><u>.</u>

5 0
2 years ago
James invest some money into an account that pays 4% compounded annually if the account is $851.66 after five years how much do
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Your moma

Step-by-step explanation:

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3 years ago
Read 2 more answers
In an origami class, the instructor tells the students to cut a square that has an area of 24 square inches. Kevin cuts a square
rodikova [14]
The answer is
The side length of the square Kevin cuts is slightly less than the length the instructor required
7 0
3 years ago
A man can drive a motorboat 70 miles down the Colorado River in the same amount of time that he can drive 40 miles upstream. Fin
pochemuha

The speed of the current is 40.34 mph approximately.

<u>SOLUTION: </u>

Given, a man can drive a motorboat 70 miles down the Colorado River in the same amount of time that he can drive 40 miles upstream.  

We have to find the speed of the current if the speed of the boat is 11 mph in still water. Now, let the speed of river be a mph.  Then, speed of boat in upstream will be a-11 mph and speed in downstream will be a+11 mph.

And, we know that, \text{ distance } =\text{ speed }\times \text{ time }

\begin{array}{l}{\text { So, for upstream } \rightarrow 40=(a-11) \times \text { time taken } \rightarrow \text { time taken }=\frac{40}{a-11}} \\\\ {\text { And for downstream } \rightarrow 70=(a+11) \times \text { time taken } \rightarrow \text { time taken }=\frac{70}{a+11}}\end{array}

We are given that, time taken for both are same. So \frac{40}{a-11}=\frac{70}{a+11}

\begin{array}{l}{\rightarrow 40(a+11)=70(a-11)} \\\\ {\rightarrow 40 a+440=70 a-770} \\\\ {\rightarrow 70 a-40 a=770+440} \\\\ {\rightarrow 30 a=1210} \\\\ {\rightarrow a=40.33}\end{array}

8 0
3 years ago
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