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Salsk061 [2.6K]
3 years ago
8

Answer this pls fast ASAP

Mathematics
2 answers:
likoan [24]3 years ago
5 0

Answer:

14.13

Step-by-step explanation:

6/2=3

3^2=9

9x3.14=28.26

28.26/2=14.13

Katarina [22]3 years ago
4 0

Answer:

14.13 units^2.

Step-by-step explanation:

The area of a circle is found by doing pi * r^2. In this case, the diameter is 6. 6 / 2 = 3. So, r = 3.

pi * 3^2 = pi * 9 = 9pi.

This is a semicircle, so the area will be half of the whole circle.

9pi / 2 = 4.5 * pi = 4.5 * 3.14 = 14.13 units^2.

Hope this helps!

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Answer:

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3 years ago
3x/5 - x = x/5 - 5/2?
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X= 4.167 you cross multiply both the equations. so you get 30x - 50x = 10x - 125. you shift all the values of x on side and then divide it by 125 sso you get x= 4.167 
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3 years ago
Given: segment AB || segment DE, C is the midpoint of segment DB. Prove: ΔACB ≅ ΔECD Fill in the missing reason for the proof. A
OleMash [197]

Answer:

Given: Segment AB || segment DE, C is the midpoint of segment DB.

Prove: ΔA CB ≅ ΔE CD

Proof: In ΔA CB and ΔE CD

 C is the Mid point of B D.

BC=C D→ definition of midpoint

∠A CB= ∠ EC D→→vertical angles are congruent

∠BAC=∠DEC→→[AB║DE,so alternate angles are equal]

→→ΔA CB ≅ ΔE CD[A AS or A SA]

Option B: vertical angles are congruent

4 0
3 years ago
Read 2 more answers
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
Question down below thanks
son4ous [18]

Answer:

A

Thanks mate !

8 0
2 years ago
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