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Flauer [41]
3 years ago
15

The perimeter of a square in which A=169cm square root 2?

Mathematics
1 answer:
sergey [27]3 years ago
7 0
I'm guessing you mean:
A = 169 \sqrt{2}

If not, correct me. 

So, then the perimeter of a square is just 4*A, which in this case would be:
4(169 \sqrt{2}) = 676 \sqrt{2}

And that would be your answer: 676 \sqrt{2} units²
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Please help no idea how to start it <br> The radius is 2.5 m
Talja [164]

2.5 \div 2 = 1.25
so 1.25 is the right answer not sure tho
6 0
3 years ago
If x^2+y^2=1, what is the largest possible value of |x|+|y|?
Marianna [84]

If <em>x</em>² + <em>y</em>² = 1, then <em>y</em> = ±√(1 - <em>x</em>²).

Let <em>f(x)</em> = |<em>x</em>| + |±√(1 - <em>x</em>²)| = |<em>x</em>| + √(1 - <em>x</em>²).

If <em>x</em> < 0, we have |<em>x</em>| = -<em>x</em> ; otherwise, if <em>x</em> ≥ 0, then |<em>x</em>| = <em>x</em>.

• Case 1: suppose <em>x</em> < 0. Then

<em>f(x)</em> = -<em>x</em> + √(1 - <em>x</em>²)

<em>f'(x)</em> = -1 - <em>x</em>/√(1 - <em>x</em>²) = 0   →   <em>x</em> = -1/√2   →   <em>y</em> = ±1/√2

• Case 2: suppose <em>x</em> ≥ 0. Then

<em>f(x)</em> = <em>x</em> + √(1 - <em>x</em>²)

<em>f'(x)</em> = 1 - <em>x</em>/√(1 - <em>x</em>²) = 0   →   <em>x</em> = 1/√2   →   <em>y</em> = ±1/√2

In either case, |<em>x</em>| = |<em>y</em>| = 1/√2, so the maximum value of their sum is 2/√2 = √2.

6 0
3 years ago
Drag and drop the correct angle measures onto the diagram
svlad2 [7]

Answer:

Here you go!

Hope this helps!

7 0
3 years ago
2.<br> Which of the functions below is an inverse of the quadratic function f(x) = x2 - 3 ?
mezya [45]

Answer:

the inverse of the quadratic function f(x) = x^2 - 3  is \mathbf{f^{-1}(x)=\pm \sqrt{x+3}}

Step-by-step explanation:

We need to find inverse of the quadratic function f(x) = x^2 - 3

Let

y=x^2-3

Replace x and y

x=y^2-3

Now, we will solve for y

Adding 3 on both sides

x+3=y^2-3+3

x+3=y^2

Taking square root on both sides

\sqrt{y^2}=\sqrt{x+3}\\y=\pm \sqrt{x+3}

Now replace y with f^{-1}(x)

f^{-1}(x)=\pm \sqrt{x+3}

So, the inverse of the quadratic function f(x) = x^2 - 3  is \mathbf{f^{-1}(x)=\pm \sqrt{x+3}}

3 0
3 years ago
How do you determine if a relation is quadratic by calculation differences, analyzing a graph, examining the degree in an equati
Kryger [21]

1) We can determine by the table of values whether a function is a quadratic one by considering this example:

x | y 1st difference 2nd difference

0 0 3 -0 = 3 7-3 = 4

1 3 10 -3 = 7 11 -7 = 4

2 10 21 -10 =11 15 -11 = 4

3 21 36-21 = 15 19-5 = 4

4 36 55-36= 19

5 55

2) Let's subtract the values of y this way:

3 -0 = 3

10 -3 = 7

21 -10 = 11

36 -21 = 15

55 -36 = 19

Now let's subtract the differences we've just found:

7 -3 = 4

11-7 = 4

15-11 = 4

19-15 = 4

So, if the "second difference" is constant (same result) then it means we have a quadratic function just by analyzing the table.

3) Hence, we can determine if this is a quadratic relation calculating the second difference of the y-values if the second difference yields the same value. The graph must be a parabola and the highest coefficient must be 2

4 0
1 year ago
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