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3 years ago

What is the equation, in point slope form, of the line

Mathematics

1 answer:

melamori03 [73]3 years ago

5 0

Answer:

y=3/2x+11/2

Step-by-step explanation:

Hello! Sorry I just saw this

Anyways, let's continue

first, we need to find the equation of the line with the one that is (-2,-4) and (2,2)

first, we need to find the slope

the equation for slope is y2-y1/x2-x1

so let's label the points

x1=-2

y1=-4

x2=2

y2=2

now plug it in

2-(-4)/2-(-2)=6/4=3/2

now, let's turn it into a line

the point-slope form is y-y1=m(x-x1) (m=slope)

now, plug it in

y-(-4)=3/2(x-(-2))

simplify to

y+4=3/2(x+2)

turn into y=mx+b format

y+4=3/2x+3

subtract 4 on both sides

y=3/2x-1

Now for the line that is parallel.

Parallel lines have the same slopes, so you automatically know that the new line will be y=3/2x+b

To make sure (-3,1) is a solution to the point, put 1 as y and -3 as x

1=3/2(-3)+b

1=-9/2+b

add 9/2 on both sides

b=11/2 or 5.5

now, put it into the equation

y=3/2x+11/2

Hope this helps!

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In cartesian plane the general equation of a line is given by :

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2 years ago

Graph the line with slope 4 and y-intercept -7

antoniya [11.8K]

Start on -7 since it is the y intercept(b). 4/1 is the slope. So from seven you would go up 4 right 1.

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3 years ago

Read 2 more answers

Which of the following is the factored form of 9y2-121

lukranit [14]

It's the first one.

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4 0

3 years ago

3. The adult men of the Dinaric Alps have the highest average height of all regions. The

ahrayia [7]

Using the normal distribution, it is found that:

3 - a) The 40th percentile of the height of Dinaric Alps distribution for men is of 72.2 inches.

3 - b) The minimum height of man in the Dinaric Alps that would place him in the top 10% of all heights is of 76.84 inches.

4 - a) The 25th percentile for the math scores was of 71.6 inches.

4 - b) The 75th percentile for the math scores was of 78.4 inches.

<h3>Normal Probability Distribution </h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

Question 3:

The mean is of 73 inches, hence $\mu = 73$ .

The standard deviation is of 3 inches, hence $\sigma = 3$ .

Item a:

The 40th percentile is X when Z has a p-value of 0.4, so X when Z = -0.253.

$Z = \frac{X - \mu}{\sigma}$

$-0.253 = \frac{X - 73}{3}$

$X - 73 = -0.253(3)$

$X = 72.2$

The 40th percentile of the height of Dinaric Alps distribution for men is of 72.2 inches.

Item b:

The minimum height is the 100 - 10 = 90th percentile is X when Z has a p-value of 0.9, so X when Z = 1.28.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 73}{3}$

$X - 73 = 1.28(3)$

$X = 76.84$

The minimum height of man in the Dinaric Alps that would place him in the top 10% of all heights is of 76.84 inches.

Question 4:

The mean score is of 75, hence $\mu = 75$ .

The standard deviation is of 5, hence $\sigma = 5$ .

Item a:

The 25th percentile is X when Z has a p-value of 0.25, so X when Z = -0.675.

$Z = \frac{X - \mu}{\sigma}$

$-0.675 = \frac{X - 75}{5}$

$X - 75 = -0.675(5)$

$X = 71.6$

The 25th percentile for the math scores was of 71.6 inches.

Item b:

The 75th percentile is X when Z has a p-value of 0.25, so X when Z = 0.675.

$Z = \frac{X - \mu}{\sigma}$

$0.675 = \frac{X - 75}{5}$

$X - 75 = 0.675(5)$

$X = 78.4$

The 75th percentile for the math scores was of 78.4 inches.

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2 years ago

Given points A(-1, -2) and B(2, 4) where AP: BP=1:2, find the locus of point P.

koban [17]

Answer:

$x^2 + 4x + y^2 +8y = 0$

Step-by-step explanation:

Given

$A = (-1,-2)$

$B = (2,4)$

$AP:BP = 1 : 2$

Required

The locus of P

$AP:BP = 1 : 2$

Express as fraction

$\frac{AP}{BP} = \frac{1}{2}$

Cross multiply

$2AP = BP$

Calculate AP and BP using the following distance formula:

$d = \sqrt{(x - x_1)^2 + (y - y_1)^2}$

So, we have:

$2 * \sqrt{(x - -1)^2 + (y - -2)^2} = \sqrt{(x - 2)^2 + (y - 4)^2}$

$2 * \sqrt{(x +1)^2 + (y +2)^2} = \sqrt{(x - 2)^2 + (y - 4)^2}$

Take square of both sides

$4 * [(x +1)^2 + (y +2)^2] = (x - 2)^2 + (y - 4)^2$

Evaluate all squares

$4 * [x^2 + 2x + 1 + y^2 +4y + 4] = x^2 - 4x + 4 + y^2 - 8y + 16$

Collect and evaluate like terms

$4 * [x^2 + 2x + y^2 +4y + 5] = x^2 - 4x + y^2 - 8y + 20$

Open brackets

$4x^2 + 8x + 4y^2 +16y + 20 = x^2 - 4x + y^2 - 8y + 20$

Collect like terms

$4x^2 - x^2 + 8x + 4x + 4y^2 -y^2 +16y + 8y + 20 - 20 = 0$

$3x^2 + 12x + 3y^2 +24y = 0$

Divide through by 3

$x^2 + 4x + y^2 +8y = 0$

3 0

3 years ago