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evablogger [386]
3 years ago
5

Answer iiuhchhgfhhvccc

Mathematics
1 answer:
ICE Princess25 [194]3 years ago
5 0
Combine ln 24 - (1/2)ln 9:  That comes out to  ln 24 - ln sqrt(9) = ln 24 - ln 3, which in turn is equal to ln 24/3, or ln 8.

Then ln (k^2 - 1) = ln 8, and  so  k^2 - 1 = 8,   k^2   = 9   and 

k = plus or minus 3. 
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A circle has the equation 2x2+12x+2y2−16y−150=0.
astra-53 [7]

Answer:

Step-by-step explanation:

hello :

2x²+12x+2y²−16y−150=0

divide by 2 : (x²+6x)+(y²-8y)-75 =0

(x²+6x+9)-9+(y²-8y+16)-16-75 =0

(x+3)²+(y-4)² =10²

The coordinates of the center are (−3,4), and the length of the radius is 10 units.

6 0
3 years ago
What is the mean of this set: {2, 6, 7, 9, 9, 9)?<br> 6<br> 7<br> Ο Ο Ο Ο<br> 8<br> 9
vovikov84 [41]

Answer:

6.8 nearly 7

Step-by-step explanation:

beacuse mean is sum of all terms in a data/number of terms in the data

3 0
3 years ago
Read 2 more answers
The equation of function h is h... PLEASE HELP MATH
Flura [38]

Answer:

Part A: the value of h(4) - m(16) is -4

Part B: The y-intercepts are 4 units apart

Part C: m(x) can not exceed h(x) for any value of x

Step-by-step explanation:

Let us use the table to find the function m(x)

There is a constant difference between each two consecutive values of x and also in y, then the table represents a linear function

The form of the linear function is m(x) = a x + b, where

  • a is the slope of the function
  • b is the y-intercept

The slope = Δm(x)/Δx

∵ At x = 8, m(x) = 2

∵ At x = 10, m(x) = 3

∴ The slope = \frac{3-2}{10-8}=\frac{1}{2}

∴ a = \frac{1}{2}

- Substitute it in the form of the function

∴ m(x) = \frac{1}{2} x + b

- To find b substitute x and m(x) in the function by (8 , 2)

∵ 2 = \frac{1}{2} (8) + b

∴ 2 = 4 + b

- Subtract 4 from both sides

∴ -2 = b

∴ m(x) = \frac{1}{2} x - 2

Now let us answer the questions

Part A:

∵ h(x) = \frac{1}{2} (x - 2)²

∴ h(4) = \frac{1}{2} (4 - 2)²

∴ h(4) = \frac{1}{2} (2)²

∴ h(4) =  \frac{1}{2}(4)

∴ h(4) = 2

∵ m(x) = \frac{1}{2} x - 2

∴ m(16) =  \frac{1}{2} (16) - 2

∴ m(16) = 8 - 2

∴ m(16) = 6

- Find now h(4) - m(16)

∵ h(4) - m(16) = 2 - 6

∴ h(4) - m(16) = -4

Part B:

The y-intercept is the value of h(x) at x = 0

∵ h(x) = \frac{1}{2} (x - 2)²

∵ x = 0

∴ h(0) = \frac{1}{2} (0 - 2)²

∴ h(0) =  \frac{1}{2} (-2)² =  

∴ h(0) = 2

∴ The y-intercept of h(x) is 2

∵ m(x) = \frac{1}{2} x - 2

∵ x = 0

∴ m(0) = \frac{1}{2} (0) - 2 = 0 - 2

∴ m(0) = -2

∴ The y-intercept of m(x) is -2

- Find the distance between y = 2 and y = -2

∴ The difference between the y-intercepts of the graphs = 2 - (-2)

∴ The difference between the y-intercepts of the graphs = 4

∴ The y-intercepts are 4 units apart

Part C:

The minimum/maximum point of a quadratic function f(x) = a(x - h) + k is point (h , k)

Compare this form with the form of h(x)

∵ h = 2 and k = 0

∴ The minimum point of the graph of h(x) is (2 , 0)

∵ k is the minimum value of f(x)

∴ 0 is the minimum value of h(x)

∴ The domain of h(x) is all real numbers

∴ The range of h(x) is h(x) ≥ 2

∵ m(8) = 2

∵ m(14) = 5

∵ h(8) = \frac{1}{2} (8 - 2)² = 18

∵ h(14) = \frac{1}{2} (14 - 2)² = 72

∴ h(x) is always > m(x)

∴ m(x) can not exceed h(x) for any value of x

<em>Look to the attached graph for more understand</em>

The blue graph represents h(x)

The green graph represents m(x)

The blue graph is above the green graph for all values of x, then there is no value of x make m(x) exceeds h(x)

7 0
3 years ago
NO LINKS!!!
Ber [7]

Refer to this previous solution set

brainly.com/question/26114608

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

Problem 4

Like the three earlier problems, we'll place the kicker at the origin and have her kick to the right. The two roots in this case are x = 0 and x = 20 to represent when the ball is on the ground.

This leads to the factors x and x-20 and the equation y = ax(x-20)

We'll plug in (x,y) = (10,28) which is the vertex point. The 10 is the midpoint of 0 and 20 mentioned earlier.

Let's solve for 'a'.

y = ax(x-20)\\\\28 = a*10(10-20)\\\\28 = -100a\\\\a = -\frac{28}{100}\\\\a = -\frac{7}{25}\\\\

This then leads us to:

y = ax(x-20)\\\\y = -\frac{7}{25}x(x-20)\\\\y = -\frac{7}{25}x*x-\frac{7}{25}x*(-20)\\\\y = -\frac{7}{25}x^2+\frac{28}{5}x\\\\

The equation is in the form y = ax^2+bx+c with a = -\frac{7}{25}, \ b = \frac{28}{5}, \ c = 0

The graph is below in blue.

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

Problem 5

The same set up applies as before.

This time we have the roots x = 0 and x = 100 to lead to the factors x and x-100. We have the equation y = ax(x-100)

We'll use the vertex point (50,12) to find 'a'.

y = ax(x-100)\\\\12 = a*50(50-100)\\\\12 = -2500a\\\\a = -\frac{12}{2500}\\\\a = -\frac{3}{625}\\\\

Then we can find the standard form

y = ax(x-100)\\\\y = -\frac{3}{625}x(x-100)\\\\y = -\frac{3}{625}x*x-\frac{3}{625}x*(-100)\\\\y = -\frac{3}{625}x^2+\frac{12}{25}x\\\\

The graph is below in red.

4 0
2 years ago
Domain and range of (4,-4),(2,-3),(-5,1),(4,-1)
aniked [119]

(x; y)


The domain is x.

The range is y.


Therefore

(4; -4) → x = 4; y = -4

(2;-3) → x = 2; y = -3

(-5; 1) → x = -5; y = 1

if is (-4; -1) not (4; -1) then

(-4; -1) → x = -4; y = -1


The DOMAIN = {-5; -4; 2; 4}

The RANGE = {-4; -3; -1; 1}


If is (4; -4) and (4; -1) then it is not a function.

4 0
3 years ago
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