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klemol [59]
3 years ago
9

Help me I need to pass so I can go on to the next thing plz someone help me. Will give brainliest!!​

Mathematics
1 answer:
expeople1 [14]3 years ago
5 0
I think its 7 or 10 Dont remember that question
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Student Council sponsors weekly dances at their school on Friday nights. The admission price for each person is $4 for Student C
Morgarella [4.7K]

Answer:

$110

Step-by-step explanation:

So first you have to do 15*4. We get the 4 from the admission fees and the 15 from the number of dances they attend. From this answer we can see the total of money they will be paying for admission for the 15 dances.

15*4=$60

However, our answer isn't just $50 as we are also told membership costs $50. So no we add $50 to $60, which gives us an answer of $110, which is the amount the member will pay if they attend 15 dances during the school year.

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3 years ago
Gggtfttttttttttttttttttttttt
Marizza181 [45]

Answer:

A) You would move it 2 times B) The Exponent would be 10^2

Step-by-step explanation:

This one's hard to explain..if you want you can message me about it :)

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3 years ago
What number is 250% of 4.2?​
agasfer [191]

Answer

10.5

Step-by-step explanation:

6 0
3 years ago
What is 13x + 21y - 14x - y simplified?
Arisa [49]

Answer:

-x + 20y

Step-by-step explanation:

trust me.. its right

4 0
3 years ago
Suppose that f : [a, b] → [a, b] is continuous. Prove that f has a fixed point. That is, prove that there exists c ∈ [a, b] such
nikklg [1K]

Answer:

Step-by-step explanation:

define the function:

g(x) = f(x) -x

As both f(x) and x are continuous functions, g(x) will also be continuous.

Now, what can we say about g(a) = f(a) -a?

we know that a\leq f(a) \leq b, thus:

a-a\leq f(a)-a \leq b-a\\0 \leq g(a) \leq b-a

thus  g(a) is non-negative.

What about g(b) ? Again we have:

a\leq f(b) \leq b\\a-b \leq f(b) -b \leq  0\\a-b \leq g(b) \leq  0

That means that g(b) is not positive.

Now, we can imagine two cases, either one of g(a) or g(b) is equal to zero, or none of them is. If either of them is equal to zero, we have found a fixed point! In fact, any point c for which g(c)=0 is a fixed point, because:

g(c) = 0 \implies f(c) -c = 0 \implies f(c) = c

Now, if g(a) \neq  0 and g(b) \neq 0, then we have that

g(a) >0 and g(b) < 0. And by Bolzano's theorem we can assert that there must exist a point c between a and b for which g(c)=0. And as we have shown before that point would be a fixed point. This completes the proof.

6 0
3 years ago
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