Ask question

Ask question

All categories

valentina_108 [34]

3 years ago

8

How do I do this? Please help

1 answer:

Murljashka [212]3 years ago

5 0

X is 90 degrees. If AB is the diameter, then it passes through the centre. X is 90 degrees because the angle in a semicircle is always a right angle.

You might be interested in

The expression has exactly 2 terms terms true or false 7f + 15 - 2g

Dvinal [7]

In this case, the answer is very simple.

A term is a part not separated by the signs + or - .

In this expression, we have 3 terms.

The answer is: false.

3 0

1 year ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%2096" id="TexFormula1" title="\sqrt 96" alt="\sqrt 96" align="absmiddle" class="latex-

klasskru [66]

Answer:

If it is $\sqrt{96} \sqrt{96}$ then it is the solution $\sqrt{96} \sqrt{96} = 96$ because the square root of an expression multiplied by itself gives that expression.

7 0

2 years ago

Show some work please or an explanation please. I'll give out brainlest.<br><br> 3x + 5 = -x -7

Viefleur [7K]

Step-by-step explanation:

3x+x=-7-5

4x=-12

x=-12/4

x=-3

7 0

3 years ago

Read 2 more answers

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!! Will mark brainilest, thank you in advance. :))

qaws [65]

Answer:

See image below:) :)

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers