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2 years ago

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Decide whether the table could represent a proportional relationship. If the relationship could be proportional, what would be t

he constant of proportionality?

2 answers:

Savatey [412]2 years ago

7 0

Non proportional rate 3

Feliz [49]2 years ago

5 0

Answer:

a) non-proportional/ constant rate: 3

b) non-proportional/ constant rate 3

Step-by-step explanation:

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If the rate of change of the graph of a linear equation is 4 and the y-intercept is (0, 8), what is the linear equation that rep

serg [7]

Answer:

y = 4x + 8

Step-by-step explanation:

The linear equation can be written in the form y = mx + b. "y" and "x" stay as they are. The "m" is the slope or rate of change. The b is the y-intercept.

Hope it helps!!!

(May I have brainliest if it's correct please?)

7 0

2 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: <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>

7 0

3 years ago

7900 dollars is placed in an account with an annual interest rate of 5.5%. How much will be in the account after 11 years, to th

Lady bird [3.3K]

Answer:

$\$12,679.50$

Step-by-step explanation:

we know that

The simple interest formula is equal to

$A=P(1+rt)$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest

t is Number of Time Periods

in this problem we have

$t=11\ years\\ P=\$7,900\\r=5.5\%=5.5/100=0.055$

substitute in the formula above

$A=7,900(1+0.055*11)$

$A=7,900(1.605)$

$A=\$12,679.50$

4 0

3 years ago

Read 2 more answers

For geometry:(<br><br> i’ll give a brainist!!!

kvasek [131]

<h3>Answer: Angle Q = 133 degrees</h3>

========================================================

Work Shown:

Recall that for any triangle, the three angles always add to 180

P+Q+R = 180

(x+13) + (10x+13) + (2x-2) = 180

(x+10x+2x) + (13+13-2) = 180

13x+24 = 180

13x = 180-24

13x = 156

x = 156/13

x = 12

Now that we know x, we can find the angle measures

angle P = x+13 = 12+13 = 25 degrees
angle Q = 10x+13 = 10*12+13 = 120+13 = 133 degrees
angle R = 2x-2 = 2*12-2 = 24-2 = 22 degrees

As a way to check if we have the right answer or not, we see that,

P+Q+R = 25+133+22 = 180

So the answer is confirmed.

4 0

2 years ago

The equation that describes the relationship is y =

Rainbow [258]

I just need point tbh so thanks !!

7 0

2 years ago

Read 2 more answers