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amid [387]
3 years ago
15

Find the inverse of y=100-x^2

Mathematics
1 answer:
AnnyKZ [126]3 years ago
5 0

Answer:

The inverse  is ±sqrt(100-x)

Step-by-step explanation:

To find the inverse, exchange x and y

x = 100 -y^2

Solve for y

Subtract 100 from each side

x - 100 = -y^2

Divide by -1

-x +100 = y^2

Take the square root of each side

±sqrt(100-x) = sqrt(y^2)

±sqrt(100-x) =y

The inverse  is ±sqrt(100-x)

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John and alisha randomly met at the local movie theater. what will affect their first impression of each other?
Vesna [10]
How they dress, What movie they buy, and what snacks you buy
7 0
3 years ago
Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is incre
Ainat [17]

Answer: (a) Interval where f is increasing: (0.78,+∞);

Interval where f is decreasing: (0,0.78);

(b) Local minimum: (0.78, - 0.09)

(c) Inflection point: (0.56,-0.06)

Interval concave up: (0.56,+∞)

Interval concave down: (0,0.56)

Step-by-step explanation:

(a) To determine the interval where function f is increasing or decreasing, first derive the function:

f'(x) = \frac{d}{dx}[x^{4}ln(x)]

Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),

you have:

f'(x) = 4x^{3}ln(x) + x_{4}.\frac{1}{x}

f'(x) = 4x^{3}ln(x) + x^{3}

f'(x) = x^{3}[4ln(x) + 1]

Now, find the critical points: f'(x) = 0

x^{3}[4ln(x) + 1] = 0

x^{3} = 0

x = 0

and

4ln(x) + 1 = 0

ln(x) = \frac{-1}{4}

x = e^{\frac{-1}{4} }

x = 0.78

To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:

interval                 x-value                      f'(x)                       result

0<x<0.78                 0.5                 f'(0.5) = -0.22            decreasing

x>0.78                       1                         f'(1) = 1                  increasing

With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.

Note: As it is a natural logarithm function, there are no negative x-values.

(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:

  • Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
  • After 0.78, it increase (has a change of sign) and f is also defined;

Then, x=0.78 is a point of minimum and its y-value is:

f(x) = x^{4}ln(x)

f(0.78) = 0.78^{4}ln(0.78)

f(0.78) = - 0.092

The point of <u>minimum</u> is (0.78, - 0.092)

(c) To determine the inflection point (IP), calculate the second derivative of the function and solve for x:

f"(x) = \frac{d^{2}}{dx^{2}} [x^{3}[4ln(x) + 1]]

f"(x) = 3x^{2}[4ln(x) + 1] + 4x^{2}

f"(x) = x^{2}[12ln(x) + 7]

x^{2}[12ln(x) + 7] = 0

x^{2} = 0\\x = 0

and

12ln(x) + 7 = 0\\ln(x) = \frac{-7}{12} \\x = e^{\frac{-7}{12} }\\x = 0.56

Substituing x in the function:

f(x) = x^{4}ln(x)

f(0.56) = 0.56^{4} ln(0.56)

f(0.56) = - 0.06

The <u>inflection point</u> will be: (0.56, - 0.06)

In a function, the concave is down when f"(x) < 0 and up when f"(x) > 0, adn knowing that the critical points for that derivative are 0 and 0.56:

f"(x) =  x^{2}[12ln(x) + 7]

f"(0.1) = 0.1^{2}[12ln(0.1)+7]

f"(0.1) = - 0.21, i.e. <u>Concave</u> is <u>DOWN.</u>

f"(0.7) = 0.7^{2}[12ln(0.7)+7]

f"(0.7) = + 1.33, i.e. <u>Concave</u> is <u>UP.</u>

4 0
3 years ago
The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?
Nuetrik [128]

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

A(-6, 2)) = (x_1, y_1)

B(2, 6) = (x_2, y_2)

AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}

AB = \sqrt{(8)^2 + (4)^2}

AB = \sqrt{64 + 16} = \sqrt{80}

AB = 8.9 (nearest tenth)

Distance between B(2, 6) and C(7, 1):

BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

B(2, 6) = (x_1, y_1)

C(7, 1) = (x_2, y_2)

BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}

BC = \sqrt{(5)^2 + (-5)^2}

BC = \sqrt{25 + 25} = \sqrt{50}

BC = 7.1 (nearest tenth)

Distance between C(7, 1) and D(3, -5):

CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

C(7, 1) = (x_1, y_1)

D(3, -5) = (x_2, y_2)

CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}

CD = \sqrt{(-4)^2 + (-6)^2}

CD = \sqrt{16 + 36} = \sqrt{52}

CD = 7.2 (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

D(3, -5) = (x_1, y_1)

A(-6, 2) = (x_2, y_2)

DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}

DA = \sqrt{(-9)^2 + (7)^2}

DA = \sqrt{81 + 49} = \sqrt{130}

DA = 11.4 (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

8 0
3 years ago
‼️GEOMETRY‼️
ella [17]

Answer:

A.) 1018 square inches

Step-by-step explanation:

The largest sphere will have a diameter equaling the length of the cube (see picture).

If the side length of the cube is 18 inches, the diameter of the sphere is also 18 inches. Use the surface area formula for a circle:

SA=4\pi r^2

For this formula, we need the radius of the sphere. Divide the diameter by 2:

\frac{18}{2}=9

The radius is 9 inches. Plug this into the equation:

SA=4\pi *9^2

Simplify the equation:

SA=4\pi *81\\\\SA=324\pi \\\\SA=1017.87

Round the result to the nearest whole number:

1017.87 → 1018

The surface area is 1,018 inches².

:Done

Picture:

In a 2D version, we can clearly see that if the circle fits snuggly inside of the square, the diameter of a sphere is the same as the length of a side of the cube.

6 0
3 years ago
What’s the answer??????
Fofino [41]

Answer:

2x=34+56

x=90÷2

therefore x=45

6 0
3 years ago
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