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3 years ago

8

Use substitution to solve the system of linear equations. In your final answer, include all of your work.

1 answer:

Anna11 [10]3 years ago

4 0

Answer:

Y=-3

Step-by-step explanation:

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F(x)= 2x^2 + k/x point of inflection at x=-1 then value of k is

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I think it isss 2 :)))

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3 years ago

Mark as brainliest

almond37 [142]

Answer:

C) Casey is incorrect because more than one student can have the same GPA.

Step-by-step explanation:

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3 years ago

I have 2 questions and i need the answers in less than 20 mins question 1.

Debora [2.8K]

32.6 0.61
+3.54 x 0.45
--------- ----------
36.14 305
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3 years ago

Read 2 more answers

A. Use composition to prove whether or not the functions are inverses of each other. B. Express the domain of the compositions u

Kryger [21]

Given: $f(x) = \frac{1}{x-2}$

$g(x) = \frac{2x+1}{x}$

A.)Consider

$f(g(x))= f(\frac{2x+1}{x} )$

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

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

$f(\frac{2x+1}{x} )=\frac{x}{1}$

$f(\frac{2x+1}{x} )=1$

Also,

$g(f(x))= g(\frac{1}{x-2} )$

$g(\frac{1}{x-2} )= \frac{2(\frac{1}{x-2}) +1 }{\frac{1}{x-2}}$

$g(\frac{1}{x-2} )= \frac{\frac{2+x-2}{x-2} }{\frac{1}{x-2}}$

$g(\frac{1}{x-2} )= \frac{x }{1}$

$g(\frac{1}{x-2} )= x$

Since, $f(g(x))=g(f(x))=x$

Therefore, both functions are inverses of each other.

B.

For the Composition function $f(g(x)) = f(\frac{2x+1}{x} )=x$

Since, the function $f(g(x))$ is not defined for $x=0$ .

Therefore, the domain is $(-\infty,0)\cup(0,\infty)$

For the Composition function $g(f(x)) =g(\frac{1}{x-2} )=x$

Since, the function $g(f(x))$ is not defined for $x=2$ .

Therefore, the domain is $(-\infty,2)\cup(2,\infty)$

8 0

3 years ago

What is the value of m so that the two expressions are equivalent?

Law Incorporation [45]

Answer: 0

Step-by-step explanation:

Since m is with x, you don't need to care about y

We would get 2x - 6x + 4x = mx

0x = mx

m is 0

8 0

2 years ago