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3 years ago

I am so confused on functions! Please help!

1 answer:

4 0

Functions are basically another way of expressing a way to get y or x or anything (mostly used for graphs) You will see functions like f(x)= 23y-5 (that was made up). Basically the number in the parenthesis in front of the f is what you're finding. So the equation after the equal sign is how to find whatever is being found. So in this case, finding x can be pretty easy. By the way, when graphing this you have x as the run and y as the rise.

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the company president sets a goal that the percentage of working phones must increase from 30% to at least 80% by the end of the

MArishka [77]

Answer:

Step-by-step explanation:

5 0

3 years ago

Point C is the Image of C (-4,-2) under a Reflection across the x-axis.

Svetradugi [14.3K]

Step-by-step explanation:

$\text {Reflection Rule for across the X-axis: } (x,y) \rightarrow (x,-y)$

Using the point C (-4,-2) and the refection rule:

$(-4,-2) \rightarrow(-4, 2)$

$\text {C' should be (-4,2)}$

8 0

3 years ago

Solve this please .......

nikdorinn [45]

Answer:

The sum of reciprocals is 2/3.

You don't need complex numbers to solve this, but if you try to find a and b you will need complex numbers.

Step-by-step explanation:

a+b = 2

a*b = 3

1/a + 1/b = x

(a*b)*(1/a + 1/b) = (a*b)x

b + a = (a*b)(x)

2 = 3x

x = 2/3

b = 2 - a

a*(2 - a) = 3

-a^2 + 2a = 3

-a^2 + 2a - 3 = 0

a^2 - 2a + 3 = 0

let's solve the quadratic equation

a^2 - 2a + 3 = a^2 - 2a + 1 + 2 = (a - 1)^2 + 2 = 0

(a - 1)^2 = -2

$a_1 - 1 = \sqrt{2}i\\a_1 = 1+\sqrt{2}i\\\\a_2 - 1 = -\sqrt{2}i\\a_2 = 1 - \sqrt{2}i\\$

these options correspond to a and b from the original question.

8 0

2 years ago

Consider the functions f and g defined by \[f(x) = \sqrt{\dfrac{x+1}{x-1}}\qquad\qquad\text{and}\qquad\qquad g(x) = \dfrac{\sqrt

tino4ka555 [31]

Answer:

The given functions are not same because the domain of both functions are different.

Step-by-step explanation:

The given functions are

$f(x)= \sqrt{\dfrac{x+1}{x-1}}$

$g(x) = \dfrac{\sqrt{x+1}}{\sqrt{x-1}}$

First find the domain of both functions. Radicand can not be negative.

Domain of f(x):

$\dfrac{x+1}{x-1}>0$

This is possible if both numerator or denominator are either positive or negative.

Case 1: Both numerator or denominator are positive.

$x+1\geq 0\Rightarrow x\geq -1$

$x-1\geq 0\Rightarrow x\geq 1$

So, the function is defined for x≥1.

Case 2: Both numerator or denominator are negative.

$x+1\leq 0\Rightarrow x\leq -1$

$x-1\leq 0\Rightarrow x\leq 1$

So, the function is defined for x≤-1.

From case 1 and 2 the domain of the function f(x) is (-∞,-1]∪[1,∞).

Domain of g(x):

$x+1\geq 0\Rightarrow x\geq -1$

$x-1\geq 0\Rightarrow x\geq 1$

So, the function is defined for x≥1.

So, domain of g(x) is [1,∞).

Therefore, the given functions are not same because the domain of both functions are different.

4 0

3 years ago

TV=3x-24 and VX=2x+1. What is the value of VX? 1. 25 2. 99 3. 5 4. 51

Maslowich

ΔTXV is an isosceles triangle because ∡T = ∡X therefore TV = VX.

3x - 24 = 2x + 1 |subtract 2x from both sides

x - 24 = 1 |add 24 to both sides

x = 25

7 0

3 years ago

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