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andrezito [222]
3 years ago
15

Multiply. 0.35 × 0.003 =

Mathematics
1 answer:
Delicious77 [7]3 years ago
5 0
0.00105. or 21/20000,1.05x10^-3
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Need to round 402 to nearest ten and hundred
antoniya [11.8K]
The 0 in the number 402 is in the ten place. The number after it is 2. 2 is less than 5.

402 rounded to the nearest ten is 400

The 4 in the number 402 is in the hundreds place. The number after it is 0. 0 is less than 5.

402 rounded to the nearest hundred is 400
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3 years ago
Solve the system or linear equations
Vanyuwa [196]

Answer:

37. {-1, -1}.

Step-by-step explanation:

I'll solve the first one . The other can be solved in a similar way. We can use the method of elimination.

x1 - x2 = 0

3x1 - 2x2 = -1

We can multiply the first equation by -2. We then have an equation containing + 2x2 so when we add this to the second equation the 2x2 will be eliminated

So the first equation becomes:

-2x1 + 2x2 = 0     Bring down the second equation:

3x1 - 2x2 = -1       Now adding, we get:

x1 + 0 = -1

so x1 = -1.

Now we substitute this value of x1 in the original first equation:

-1 - x2 = 0

-1 = x2

x2 = -1.

So the solution set is {-1, -1}.

If there are more than 2 equations you can use a combination of substitutions and eliminations.

8 0
2 years ago
An English class consists of twenty students, and four are to be chosen to give speeches in a school competition.
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Shawn is collecting money from students at his school for charity the table gives the ratio of the number of students who have d
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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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