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3 years ago

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

1 answer:

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 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 x² + y² = 1, then y = ±√(1 - x²).

Let f(x) = |x| + |±√(1 - x²)| = |x| + √(1 - x²).

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

• Case 1: suppose x < 0. Then

f(x) = -x + √(1 - x²)

f'(x) = -1 - x/√(1 - x²) = 0 → x = -1/√2 → y = ±1/√2

• Case 2: suppose x ≥ 0. Then

f(x) = x + √(1 - x²)

f'(x) = 1 - x/√(1 - x²) = 0 → x = 1/√2 → y = ±1/√2

In either case, |x| = |y| = 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. 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