All categories

3 years ago

Why does the equation y=3x+5 Not represent a proportional relationship?

Mathematics

1 answer:

Afina-wow [57]3 years ago

6 0

Answer:

see the explanation

Step-by-step explanation:

we now that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin

In this problem we have

$y=3x+5$

This is the equation of the line in slope intercept form

where

$m=3\\b=5$

The given equation not represent a proportional relationship, because the line not pass through the origin

In a proportional relationship the value of b (y-intercept) is equal to zero

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Find the volume of the cylinder in terms of pi. The diagrams are not drawn to scale. Radius=1.7m length=15 m

inna [77]

V= π r 2 h
V= π•1.7•2•15= 160.22m²

5 0

3 years ago

Dy/dx = 2xy^2 and y(-1) = 2 find y(2)

Anarel [89]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2887301

—————

Solve the initial value problem:

dy
——— = 2xy², y = 2, when x = – 1.
dx

Separate the variables in the equation above:

$\mathsf{\dfrac{dy}{y^2}=2x\,dx}\\\\
\mathsf{y^{-2}\,dy=2x\,dx}$

Integrate both sides:

$\mathsf{\displaystyle\int\!y^{-2}\,dy=\int\!2x\,dx}\\\\\\
\mathsf{\dfrac{y^{-2+1}}{-2+1}=2\cdot \dfrac{x^{1+1}}{1+1}+C_1}\\\\\\
\mathsf{\dfrac{y^{-1}}{-1}=\diagup\hspace{-7}2\cdot \dfrac{x^2}{\diagup\hspace{-7}2}+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{y}=x^2+C_1}$

$\mathsf{\dfrac{1}{y}=-(x^2+C_1)}$

Take the reciprocal of both sides, and then you have

$\mathsf{y=-\,\dfrac{1}{x^2+C_1}\qquad\qquad where~C_1~is~a~constant\qquad (i)}$

In order to find the value of C₁ , just plug in the equation above those known values for x and y, then solve it for C₁:

y = 2, when x = – 1. So,

$\mathsf{2=-\,\dfrac{1}{1^2+C_1}}\\\\\\
\mathsf{2=-\,\dfrac{1}{1+C_1}}\\\\\\
\mathsf{-\,\dfrac{1}{2}=1+C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-1=C_1}\\\\\\
\mathsf{-\,\dfrac{1}{2}-\dfrac{2}{2}=C_1}$

$\mathsf{C_1=-\,\dfrac{3}{2}}$

Substitute that for C₁ into (i), and you have

$\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}}\\\\\\
\mathsf{y=-\,\dfrac{1}{x^2-\frac{3}{2}}\cdot \dfrac{2}{2}}\\\\\\
\mathsf{y=-\,\dfrac{2}{2x^2-3}}$

So y(– 2) is

$\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot (-2)^2-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{2\cdot 4-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{8-3}}\\\\\\
\mathsf{y\big|_{x=-2}=-\,\dfrac{2}{5}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: ordinary differential equation ode integration separable variables initial value problem differential integral calculus

7 0

3 years ago

PLEASE HELP WILL GIVE BRAINLIEST the number of students at north middle school is 1240 and is decreasing by 25 students per year

Nikolay [14]

Answer:

-88

Step-by-step explanation:

Let N(x) be the number of students at NMS for a certain number of years.

Let S(x) be the number of students at SMS for a certain number of years.

Let x be the number of years.

N(x)=1240-25x

S(x)=800-30x

When # of students is equal, N(x)=S(x). Therefore, we are looking for the value of x (# of years) when 1240-25x=800-30x

Subtract 1240 from each side

-25x=-440-30x

Add 30x

5x=-400

x=-88 years

Check your numbers.

6 0

3 years ago

How many vertical asymptotes does the graph of this function have?F(x)=2/x(x+1)

Fantom [35]

Answer:

Step-by-step explanation:

It has one asymptote at x = -1,0

Have a nice day!

5 0

2 years ago

I am thinking of two numbers. The sum of the numbers is 5. The first number is one less than the second number. Write a system o

Dominik [7]

Answer:

x + y = 5

y = x - 1

Hope this helped!

7 0

2 years ago