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pishuonlain [190]

3 years ago

6

Which functions domains consist of all real numbers except -2?

2 answers:

slamgirl [31]3 years ago

8 0

<span>I believe its f(x)=1/x+2. Are there choices?

</span>

katrin2010 [14]3 years ago

6 0

Is -2 the only point of reference given?

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Question 5 options: Solve for y in terms of x: 12x−13=3y+b

Elden [556K]

Y = - b/3 +4x - 13/3

8 0

3 years ago

Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.

Let's start with setting the $y$ 's equal to find those $x$ 's for which the $y$ 's are the same.

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

By power rule:

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

Since $\log(u)=\log(v)$ implies $u=v$ :

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

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

This is a quadratic equation.

Subtract $(x+1)$ on both sides:

$x^2-(x+1)=0$

$x^2-x-1=0$

Comparing this to $ax^2+bx+c=0$ we see the following:

$a=1$

$b=-1$

$c=-1$

Let's plug them into the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$x=\frac{1 \pm \sqrt{1+4}}{2}$

$x=\frac{1 \pm \sqrt{5}}{2}$

So we have the solutions to the quadratic equation are:

$x=\frac{1+\sqrt{5}}{2}$ or $x=\frac{1-\sqrt{5}}{2}$ .

The second solution definitely gives at least one of the logarithm equation problems.

Example: $\log(x)$ has problems when $x \le 0$ and so the second solution is a problem.

So the $x$ where the equations intersect is at $x=\frac{1+\sqrt{5}}{2}$ .

Let's find the $y$ -coordinate.

You may use either equation.

I choose $y=\log(x)$ .

$y=\log(\frac{1+\sqrt{5}}{2})$

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

6 0

2 years ago

Please help. I don't understand it. HELP ASAP....

Vladimir [108]

Answer:

$\large\boxed{x\leq-27}$

Step-by-step explanation:

$\dfrac{x}{-9}\geq3\qquad\text{change the signs}\\\\\dfrac{x}{9}\leq-3\qquad\text{multiply both sides by 9}\\\\9\!\!\!\!\diagup^1\cdot\dfrac{x}{9\!\!\!\!\diagup_1}\leq(-3)(9)\\\\x\leq-27$

5 0

3 years ago

I mean this. when write 1.

yawa3891 [41]

If the question is 7 x 200+50+6 it equals 1792

4 0

3 years ago

Find the student’s error in solving the following inequality.

Leto [7]

Answer:

The student should have switched the direction of the inequality sign to

get –5 > x for a final answer.

Step-by-step explanation:

When multiplying or dividing an inequality by a negative "flip" the inequality sign

The student should have switched the direction of the inequality sign to

get –5 > x for a final answer.

---------------

Note:

For a more conventional appearance the entire inequality can be flipped to

–5 < x

6 0

2 years ago