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liberstina [14]
3 years ago
8

F(x)= 2x+3/x-1, x ≠ 1, dan f^-1 adalah invers dari f

Mathematics
1 answer:
Sindrei [870]3 years ago
5 0

Answer:

The answer are

A.

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

B.

{f}^{ - 1} ( - 3) = 0

Step-by-step explanation:

To find the infers we do that

f(x) =  \frac{2 x + 3}{x - 1 }  \\ (x - 1)f(x) = 2x + 3 \\ xf(x) - f(x) = 2x + 3 \\ xf(x) - 2x = f(x) + 3 \\ x(f(x) - 2) = f(x) + 3 \\  x =  \frac{f(x) + 3}{f(x) - 2}  \\ then \:  \\  {f}^{ - 1} (x) =  \frac{x + 3}{x - 2}  \\ then \\  {f }^{ - 1} ( - 3) =  \frac{ - 3 + 3}{ - 3 + 2}  = 0

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Enter the common ratio for the geometric sequence below. 4,6,9,...
UNO [17]

Answer:

1.5

Step-by-step explanation:

6/4=1.5

9/6=1.5

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3 years ago
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vesna_86 [32]

Step-by-step explanation:

-3x + 7y = 5x + 2y (you subtract the 2y from both sides)

-3× + 5y = 5x (then add 3x to both sides)

5y = 8x (use x = - 5)

5y = 8x - 5 = -40 (both sides are divided by 5)

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6 0
3 years ago
X^2 - y^2 = 100
luda_lava [24]

Answer:

20

Step-by-step explanation:

Let's start by rewriting the second equation in terms of "x":

x+y=5

Subtract y from both sides:

x=5-y

Now, substitute "5-y" for "x" in the first equation:

(5-y)^2-y^2=100

Note that:

(a-b)^2=a^2-2ab+b^2

25-10y+y^2-y^2=100

Cancel out like terms:

25-10y=100

Subtract 25 from both sides:

-10y=75

Divide both sides by -10

y=\frac{75}{-10}=\frac{15}{-2}=-\frac{15}{2}

Now, substitute this value back into either of the equations to solve for x.

x+y=5\\x-\frac{15}{2}=5\\

Add 15/2 to both sides:

x=5+\frac{15}{2}\\x=\frac{10}{2}+\frac{15}{2}\\x=\frac{25}{2}

Now, find the difference:

x-y=\frac{25}{2}-(-\frac{15}{2})=\frac{25}{2}+\frac{15}{2}=\frac{40}{2}=20

8 0
3 years ago
140% out of what gives 56
Verizon [17]


140 / 100 * X = 56

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4 0
3 years ago
Read 2 more answers
Math again yay!...Ew math
Sliva [168]

Answer:

The graph of g(x) is wider.

Step-by-step explanation:

Parent function:

f(x)=x^2

New function:

g(x)=\left(\dfrac{1}{2}x\right)^2=\dfrac{1}{4}x^2

<u>Transformations</u>:

For a > 0

f(x)+a \implies f(x) \: \textsf{translated}\:a\:\textsf{units up}

f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}

\begin{aligned} y =a\:f(x) \implies & f(x) \: \textsf{stretched/compressed vertically by a factor of}\:a\\ & \textsf{If }a > 1 \textsf{ it is stretched by a factor of}\: a\\  & \textsf{If }0 < a < 1 \textsf{ it is compressed by a factor of}\: a\\\end{aligned}

\begin{aligned} y=f(ax) \implies & f(x) \: \textsf{stretched/compressed horizontally by a factor of} \: a\\& \textsf{If }a > 1 \textsf{ it is compressed by a factor of}\: a\\  & \textsf{If }0 < a < 1 \textsf{ it is stretched by a factor of}\: a\\\end{aligned}

If the parent function is <u>shifted ¹/₄ unit up</u>:

\implies g(x)=x^2+\dfrac{1}{4}

If the parent function is <u>shifted ¹/₄ unit down</u>:

\implies g(x)=x^2-\dfrac{1}{4}

If the parent function is <u>compressed vertically</u> by a factor of ¹/₄:

\implies g(x)=\dfrac{1}{4}x^2

If the parent function is <u>stretched horizontally</u> by a factor of ¹/₂:

\implies g(x)=\left(\dfrac{1}{2}x\right)^2

Therefore, a vertical compression and a horizontal stretch mean that the graph of g(x) is wider.

4 0
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