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mart [117]
3 years ago
11

What is the solution set for 4x + 1 = 13, given the replacement set {1, 2, 3, 4} ?

Mathematics
2 answers:
Svetradugi [14.3K]3 years ago
7 0
B-x=3

Solve for x-
4x+1-1=13-1
4x=12
4x/4=12/4
x=3
tangare [24]3 years ago
3 0
The answer is D. x = 2.
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Part 4: Use the information provided to write the vertex formula of each parabola.
sergey [27]

Answer:  1. x = (y - 2)² + 8

              \bold{2.\quad x=-\dfrac{1}{2}(y-10)^2}+1

               3. y = 2(x +9)² + 7

<u>Step-by-step explanation:</u>

Notes: Vertex form is: y =a(x - h)² + k    or      x =a(y - k)² + h

  • (h, k) is the vertex
  • point of vertex is midpoint of focus and directrix:   \dfrac{focus+directrix}{2}

     \bullet\quad a=\dfrac{1}{4p}

  • p is the distance from the vertex to the focus

1)

focus = \bigg(\dfrac{-31}{4},2\bigg)\qquad directrix: x=\dfrac{-33}{4}\\\\\text{Since directrix is x, then the x-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{-31}{4}+\frac{-33}{4}}{2}=\dfrac{\frac{-64}{4}}{2}=\dfrac{-16}{2}=-8\\\\\text{The y-value of the vertex is given by the focus as: 2}\\\\\text{vertex (h, k)}=(-8,2)

Now let's find the a-value:

p=focus-vertex\\\\p=\dfrac{-31}{4}-\dfrac{-32}{4}=\dfrac{1}{4}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{1}{4})}=\dfrac{1}{1}=1

Now, plug in a = 1   and    (h, k) = (-8, 2) into the equation x =a(y - k)² + h

x = (y - 2)² + 8

***************************************************************************************

2)

focus = \bigg(\dfrac{1}{2},10\bigg)\qquad directrix: x=\dfrac{3}{2}\\\\\text{Since directrix is x, then the x-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{1}{2}+\frac{3}{2}}{2}=\dfrac{\frac{4}{2}}{2}=\dfrac{2}{2}=1\\\\\text{The y-value of the vertex is given by the focus as: 10}\\\\\text{vertex (h, k)}=(1,10)

Now let's find the a-value:

p=focus-vertex\\\\p=\dfrac{1}{2}-\dfrac{2}{2}=\dfrac{-1}{2}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{-1}{2})}=\dfrac{1}{-2}=-\dfrac{1}{2}

Now, plug in a = -1/2   and    (h, k) = (1, 10) into the equation x =a(y - k)² + h

\bold{x=-\dfrac{1}{2}(y-10)^2}+1

***************************************************************************************

3)

focus = \bigg(-9,\dfrac{57}{8}\bigg)\qquad directrix: y=\dfrac{55}{8}\\\\\text{Since directrix is y, then the y-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{57}{8}+\frac{55}{8}}{2}=\dfrac{\frac{112}{8}}{2}=\dfrac{14}{2}=7\\\\\text{The x-value of the vertex is given by the focus as: -9}\\\\\text{vertex (h, k)}=(-9,7)

Now let's find the a-value:

p=focus-vertex\\\\p=\dfrac{57}{8}-\dfrac{56}{8}=\dfrac{1}{8}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{1}{8})}=\dfrac{1}{\frac{1}{2}}=2

Now, plug in a = 2   and    (h, k) = (-9, 7) into the equation y =a(x - h)² + k

y = 2(x +9)² + 7

4 0
3 years ago
Please help me with this I’ll give 20 points and also brainlest for whoever gave me an accurate answer and thank you!
klemol [59]

THIS ANSWER IS B TO THE ANSWER


3 0
3 years ago
Read 2 more answers
Find the taylor series for f(x) centered at the given value of a. [assume that f has a power series expansion. do not show that
Bond [772]

Taylor series is f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

To find the Taylor series for f(x) = ln(x) centering at 9, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have

f(x) = ln(x)

f^{1}(x)= \frac{1}{x} \\f^{2}(x)= -\frac{1}{x^{2} }\\f^{3}(x)= -\frac{2}{x^{3} }\\f^{4}(x)= \frac{-6}{x^{4} }

.

.

.

Since we need to have it centered at 9, we must take the value of f(9), and so on.

f(9) = ln(9)

f^{1}(9)= \frac{1}{9} \\f^{2}(9)= -\frac{1}{9^{2} }\\f^{3}(x)= -\frac{1(2)}{9^{3} }\\f^{4}(x)= \frac{-1(2)(3)}{9^{4} }

.

.

.

Following the pattern, we can see that for f^{n}(x),

f^{n}(x)=(-1)^{n-1}\frac{1.2.3.4.5...........(n-1)}{9^{n} }  \\f^{n}(x)=(-1)^{n-1}\frac{(n-1)!}{9^{n}}

This applies for n ≥ 1, Expressing f(x) in summation, we have

\sum_{n=0}^{\infinite} \frac{f^{n}(9) }{n!} (x-9)^{2}

Combining ln2 with the rest of series, we have

f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

Taylor series is f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

Find out more information about taylor series here

brainly.com/question/13057266

#SPJ4

3 0
1 year ago
Please can u answer these for me
ziro4ka [17]
18 . it's increasing by 5 every time so the next one after 23 is 28
4 0
2 years ago
How do you solve this?
maks197457 [2]
The answer is  (x-5)(x-4)
5 0
3 years ago
Read 2 more answers
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