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3 years ago

A line perpendicular to y=3x+2 and passing through the origin. Write the equation y = _____

Mathematics

1 answer:

Evgesh-ka [11]3 years ago

3 0

Answer:

y = -x / 3

Step-by-step explanation:

For this problem, it is important to note two things. We want our equation to be perpendicular to the function y = 3x + 2, and we want it to pass through the origin (0,0). With this in mind, let's begin.

To create an equation that is perpendicular to another linear function, we simply will apply the idea of a negative reciprocal to the slope of said function. Note, we are given that the function we wish to be perpendicular to has a slope of 3, as denoted in the slope-intercept form's x coefficient.

So, our function should be the negative reciprocal of this function's slope. So the negative reciprocal of 3 is -1/3. When these two slopes meet, they will form a 90-degree angle which by definition, means they are perpendicular.

The second part of this problem wants our function to go through the origin. The most simple way of satisfying this requirement is simply to make the y-intercept 0. This ensures our line will cross through the origin.

From here, let's use our information to write our equation:

y = (-1/3)x + 0

y = -x / 3

Note the bottom equation is a simplification of the top equation, where the top equation is the full slope-intercept form.

Hence, y = -x / 3 is a perpendicular line to y = 3x + 2 that also crosses at the origin.

Cheers.

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5-13)<br> -10-15<br> what’s the equation for the line that passes through points (5,-3) and (-10,15)

umka2103 [35]

Answer: y=-6/5x+3

Step-by-step explanation:

To find the equation of the line that passes through the points, you want to first find the slope by using the formula $m=\frac{y_2-y_1}{x_2-x_1}$ , then solve for the y-intercept.

$m=\frac{15-(-3)}{-10-5} =\frac{18}{-15} =-\frac{6}{5}$

Now that we have the slope, we can plug in a given point and solve for the y-intercept.

15=-6/5(-10)+b [multiply]

15=12+b [subtract both sides by 12]

b=3

Now that we have the y-intercept, we know the equation is y=-6/5x+3.

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4 years ago

winnie wrote the following riddle i am a number between 60 and 100 my ones digit is two less than my tens digits im a prime numb

Lina20 [59]

<span>The answer is that 79 IS NOT correct because the riddle says "my ones digit is 2 LESS than my tens digit.." and 79 has 7 in tens and 9 in ones which means the ones digit is 2 MORE than the tens digit.

To solve
x has to be more than 60 and less than 100
the primes between those are 61,67,71,73,79,83, 89, 97
97 is two less in the ones than in the tens so it completes all of the requirements of the riddles therefore, 97 is the answer.</span>

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4 years ago

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Emilia added 2 1⁄8 cups of brown sugar to the cookie dough and 1 2⁄8 cups of white sugar to the dough. How much more brown sugar

alexandr1967 [171]

2 1/8 - 1 2/8 = 7/8 cups more brown sugar.

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3 years ago

the marks in an examination for a Physics paper have normal distribution with mean μ and variance σ2 . 10% of the students obtai

Blababa [14]

Let $X$ be the random variable for the number of marks a given student receives on the exam.

10% of students obtain more than 75 marks, so

$P(X>75)=P\left(\dfrac{X-\mu}\sigma>\dfrac{75-\mu}\sigma\right)=P(Z>z_1)=0.10$

where $Z$ follows a standard normal distribution. The critical value for an upper-tail probability of 10% is

$P(Z>z_1)=1-F_Z(z_1)=0.10\implies z_1=F_Z^{-1}(0.90)$

where $F_Z(z)=P(Z\le z)$ denotes the CDF of $Z$ , and $F_Z^{-1}$ denotes the inverse CDF. We have

$z_1=F_Z^{-1}(0.90)\approx1.2816$

Similarly, because 20% of students obtain less than 40 marks, we have

$P(X$

so that

$P(Z$

Then $\mu,\sigma$ are such that

$\dfrac{75-\mu}\sigma\approx1.2816\implies75\approx\mu+1.2816\sigma$

$\dfrac{40-\mu}\sigma\approx-0.8416\implies40\approx\mu-0.8416\sigma$

and we find

$\mu\approx53.8739,\sigma\approx16.4848$

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4 years ago

An operations analyst counted the number of arrivals per minute at an ATM in each of 30 randomly chosen minutes. The results wer

liraira [26]

Step-by-step explanation:

since Poisson distribution parameter is not given so we have to estimate it from the sample data. The average number of of arrivals per minute at an ATM is

$\hat{\lambda}=\bar{x}=\frac{\sum x}{n}=\frac{30}{30}=1$

So probabaility for $\mathrm{X}=1$ is

$P(X=1)=\frac{e^{-\lambda} \lambda^{x}}{x !}=\frac{e^{-1} \cdot 1^{1}}{1 !}=0.3679$

So expected frequency for $X=1$ is $0.3679^{*} 30=11.037$ (or $\left.11.04\right)$ .

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3 years ago