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3 years ago

8

Name a pair of lines that apear to be parallel

1 answer:

Nitella [24]3 years ago

8 0

The sides of a square or rectangle

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The lin 2x - 3y = 4 does not passes through the point :

Ymorist [56]

Answer:

The choose D (2, –1)

2x-3y=4 —> 3y=2X-4 —> y= 2/3x – 4/3

y=mx+b —> So ; m= 2/3 , b= - 4/3

y-intercept :(0, -4/3)

0=2/3x –4/3 —> 2/3x = 4/3 —> X=2

x-intercept ; (2,0)

By drawing a straight line from point 2 on the x-axis and point -4/3 on the y-axis, the points that are on the axis have been extracted, but the point (2,-1) is not on the axis . :)

I hope it helped you ^_^

3 0

2 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

A student's grade for the four quarters are 95, 83, 91, and 85. What does the student need to earn on the final exam to have an

gulaghasi [49]

Answer:

The least score the student can get is 71 to get an overall B average at the end of the year.

Step-by-step explanation:

Let the score she needs to get be = x

We now have to take the mean of all the five subjects <em>(including x)</em> and equate it to the least score she can have to get a B grade.

Let us take the least score the student needs to get a B grade to be 85

Mean of all 5 scores =

$\frac{95+83+91+85+x}{5}=85$

$\frac{354+x}{5} =85$

$354+x = 425$

$x = 71$

The least score the student can get is 71 to get an overall B average at the end of the year.

3 0

3 years ago

A graph shows Round Target on the x-axis, from 0 to 10, and Square Target on the y-axis, from 0 to 10. A dashed straight line ha

Ray Of Light [21]

Answer:

Landa on rotundo 5 times ;....

8 0

3 years ago

Read 2 more answers

Need. help please give answer for this if its right ill mark brainiest<br> 2(5x+2)=2x+20

White raven [17]

Answer:

x=2

Step-by-step explanation:

10x+4=2x+20

10x-2x=20-4

8x=16

x=2

8 0

2 years ago

Read 2 more answers