What is the solution to the equation 1/4x−3/4=1/2, given the replacement set {−1, 1, 5} ?

Justin and George bought some baked goods. Justin purchased 5 cookies for x dollars each and 4 doughnuts for y dollars each and

borishaifa [10]

The given statement 5x=4y=12 and x+6y=5 is true.

<u>Step-by-step explanation:</u>

The question comes from linear equations in two variables.

Here, two unknown variables represent a certain situation and their value can be obtained by solving both the given equations.

The cost of cookie is represented by the variable x and cost of doughnuts can be represented by the variable y.

The coefficient of each variable in the equation represent the quantity or the number of units of each baked good.

On comparing the given equations from the data provided in the question, the statement can be stated as true.

8 0

3 years ago

Read 2 more answers

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

n 2018, homes in East Baton Rouge (EBR) Parish sold for an average of $239,000. You take a random sample of homes in Ascension p

Olenka [21]

Answer:

Conclusion

There is no sufficient evidence to conclude that the mean of the home prices from Ascension parish is higher than the EBR mean

Step-by-step explanation:

From the question we are told that

The population mean for EBR is $\mu_ 1 = \$239,000$

The sample mean for Ascension parish is $\= x_2 = \$246,000$

The p-value is $p-value = 0.045$

The level of significance is $\alpha = 0.01$

The null hypothesis is $H_o : \mu_2 = \mu_1$

The alternative hypothesis is $H_a : \mu_2 > \mu_1$

Here $\mu_2$ is the population mean for Ascension parish

From the data given values we see that

$p-value > \alpha$

So we fail to reject the null hypothesis

So we conclude that there is no sufficient evidence to conclude that the mean of the home prices from Ascension parish is higher than the EBR mean

3 0

2 years ago

Could someone help me? Thanks <3

earnstyle [38]

Answer:

D

Step-by-step explanation:

100 - 35 = 55 dollars remaining

55/20 = 3.25 hours until no money

4 0

2 years ago

Read 2 more answers

Tan (angle B) using an exact expression in trigonometric ratios in right triangles

goblinko [34]

The value of the tangent of an angle is the ratio (opposite side) / (adjacent side).

Here you see that the opposite side (opposite to Angle B) is of length 5 and the adj. side has length 12.

Thus, the tan of Angle B is 5/12. (answer)

4 0

3 years ago