Ask question

Ask question

All categories

2 years ago

12

A fence on a hill uses vertical posts L and M to hold parallel rails N and P. ∠10 and ∠14 are alternate interior angles. What is

the transversal?
A. M
B. N
C. P
D. L

1 answer:

kherson [118]2 years ago

7 0

The answer is B. Thank you:$))

You might be interested in

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

marusya05 [52]

$\large\underline{\sf{Solution-}}$

We have to find out the value of the fraction.

<u>Let us assume that:</u>

$\sf \longmapsto x =2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + ... \infty} } }$

<u>We can also write it as:</u>

$\sf \longmapsto x =2 + \dfrac{1}{x}$

$\sf \longmapsto x =\dfrac{2x + 1}{x}$

$\sf \longmapsto {x}^{2} =2x + 1$

$\sf \longmapsto {x}^{2} - 2x - 1 = 0$

<u>Comparing </u>the given <u>equation</u> with <u>ax² + bx + c = 0,</u> we get:

$\sf \longmapsto\begin{cases} \sf a =1 \\ \sf b = - 2 \\ \sf c = - 1 \end{cases}$

<u>By quadratic formula:</u>

$\sf \longmapsto x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

$\sf \longmapsto x = \dfrac{2 \pm \sqrt{ {( - 2)}^{2} - 4(1)( - 1)} }{2 \times 1}$

$\sf \longmapsto x = \dfrac{2 \pm \sqrt{4 + 4} }{2 \times 1}$

$\sf \longmapsto x = \dfrac{2 \pm \sqrt{8} }{2}$

$\sf \longmapsto x = \dfrac{2 \pm2 \sqrt{2} }{2}$

$\sf \longmapsto x = 1 \pm\sqrt{2}$

$\sf \longmapsto x = \begin{cases} \sf 1 + \sqrt{2} \\ \sf 1 - \sqrt{2} \end{cases}$

<u>But </u><u>"</u><u>x"</u><u> cannot be negative. Therefore:</u>

$\sf :\implies x = 1 + \sqrt{2}$

So, the value of the fraction is 1 + √2.

4 0

2 years ago

A class of 33 students elected a class treasurer. there were 111votes for candidate a and 15 votes for candidate

Advocard [28]

Since 111 is greater than 33, I will assume that you meant 11 votes for a candidate. Also, assuming that each student can vote only once, the answer would be

11/33 x 100=33.333333...

About 33.333 of the voters voted for candidate a.

3 0

2 years ago

Plsss help!!!!!!! #3 #4

AVprozaik [17]

Answer:

3) 162 ft² || 4) 38.25 mm²

Step-by-step explanation:

3)

area of triangle = 0.5 * base * height

= 0.5 * 27 * 12

= 162 ft²

4) area of triangle = 0.5 * base * height

= 0.5 * 8.5 * 9

= 38.25 mm²

3 0

2 years ago

Read 2 more answers

What is slope????????

Damm [24]

Step-by-step explanation:

slope another name for gradient

8 0

2 years ago

Read 2 more answers