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Kaylis [27]
3 years ago
11

Hi everyone! I have a question!

Mathematics
2 answers:
ahrayia [7]3 years ago
7 0

Answer:

Step-by-step explanation:

area of unshaded circle:

A=π*r^2=π*1=π

we have 2 circles so the area of both circles are 2π

area of the large circle is A(l)=Π*3^2=Π*9=9Π

area of the shaded region: A(s)=9Π-2Π=7Π

Tju [1.3M]3 years ago
3 0

Answer:

7π

Step-by-step explanation:

SO first find the area of the whole circle

which has a radius of 3

a=πr^2

So 3^2=9

9π

Now the two circles which are the same in area due to having the same radius

so if you plug in the 1 for radius the answer would just be π or 1π

now multiply that by 2 becuase there are 2 identical circles

π*2=2π

Now subtract

9π-2π=7π

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QUESTION TWO Determine whether or not a constant function can be
Sveta_85 [38]
<h3>Answers:</h3>
  • (a) It is <u>never</u> one-to-one
  • (b) It is <u>never</u> onto

=========================================================

Explanation:

The graph of any constant function is a horizontal flat line. The output is the same regardless of whatever input you select. Recall that a one-to-one function must pass the horizontal line test. Horizontal lines themselves fail this test. So this is sufficient to show we don't have a one-to-one function here.

Put another way: Let f(x) be a constant function. Let's say its output is 5. So f(x) = 5 no matter what you pick for x. We can then show that f(a) = f(b) = 5 where a,b are different values. This criteria is enough to show that f(x) is not one-to-one. A one-to-one function must have f(a) = f(b) lead directly to a = b. We cannot have a,b as different values.

----------------------------

The term "onto" in math, specifically when it concerns functions, refers to the idea of the entire range being accessible. If the range is the set of all real numbers, then we consider the function be onto. There's a bit more nuance, but this is the general idea.

With constant functions, we can only reach one output value (in that example above, it was the output 5). We fall very short of the goal of reaching all real numbers in the range. Therefore, this constant function and any constant function can never be onto.

4 0
2 years ago
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umka21 [38]

Answer:

c) 82

d) - 11

g)27

h)1000

k)11

l) 8

Step-by-step explanation:

{9}^{2}  + 1  \\  = (9 \times 9) + 1 \\  = 81 + 1 \\  = 82

{6}^{2}  -  {5}^{2}  \\  = (6 \times 6) - (5 \times 5) \\  = 36 - 25 \\  =  - 11

(1 + 2 {)}^{3}  \\  =  {3}^{3}  \\  = 3 \times 3  \times 3\\  = 27

(2 + 8 {)}^{3}  \\  =  {10}^{3}  \\  = (10 \times 10 \times 10) \\  = 1000

\sqrt{4}  +  {3}^{2}  \\  = 2 +  {3}^{2}  \\  = 2 + (3 \times 3) \\  = 2 + 9 \\  = 11

2 \times 5 -  \sqrt{4}  \\  = 10 -  \sqrt{4}  \\  = 10 - 2 \\  = 8

<h3>Hope it is helpful....</h3>
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