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3 years ago

Solve for x under the assumption that x<0. x-5/x<4

1 answer:

3 0

If you would like to solve the inequality x - 5/x < 4, you can do this using the following steps:

x - 5/x < 4
x^2 - 5 < 4x
x^2 - 4x - 5 < 0
(x - 5) * (x + 1) < 0
x < 0
1. x = 5
2. x = -1

The correct result would be x = -1.

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Please help me with this math problem

goldfiish [28.3K]

If you look at the graph you can tell the graph is increasing before x = -2.

From x = -2 to x = 0, it's decreasing.

Then it's increasing again from x = 0 to x = 2, then decreasing after x = 2

So answer is the last one
It is increasing before x = -2 and from x = 0 to x = 2

3 0

3 years ago

Given a quadratic function in vertex form:

Alja [10]

Answer:

See below

Step-by-step explanation:

Part A

A vertical stretch takes place when $a\cdot f(x)$ given that $|a| > 1$

Part B

A vertical compression takes place when $a\cdot f(x)$ given that $0 < |a| < 1$

Part C

A vertical stretch is different than a horizontal compression:

In a vertical stretch, the input stays the same, but the output is multiplied by the scale factor
In a horizontal compression, the output stays the same, but the input is multiplied by the scale factor

Part D

A reflection across the x-axis means that the output, our y-variable, is the opposite sign. This means that all values of $a$ must be negative such that $a\cdot f(x)$ as mentioned in parts A and B. Also, in part C, since our scale factor is negative, the output is the only one being multiplied by the scale factor.

7 0

2 years ago

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.

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

Help

mars1129 [50]

To find one year, here's the equation:
5000 + 0.06(5000)
For 10 years:
5000 + 10(0.06(5000))
Multiply:
5000 + 0.6(5000)
We can make it smaller:
1.6(5000) = 8000
You can make $8000

7 0

3 years ago

How would the expression x3 +8 be rewritten using Sum of Cubes?

Ipatiy [6.2K]

Answer:

alr so ques is

x3 + 8

x^3 + 2^3

now apply identity of a3 + b3

so finally we get

=(x+2)(x2−2x+4)

option 2

enjoy!!!!!

8 0

2 years ago