What is the correct answer someone pls help!?????

a water station is located at the midpoint of ab at what marker on the path is the water station located?

Finger [1]

Answer: Halfway between A and B.

3 0

3 years ago

Using the factorised trinomial <img src="https://tex.z-dn.net/?f=%28n-2%29%284n-7%29" id="TexFormula1" title="(n-2)(4n-7)" alt="

andrew-mc [135]

4n² - 15n + 14 is always the product of two numbers, for it to be prime number, one of these factors must be either 1 or -1.

Case n - 2 = 1
That would be n = 3
Then 4n² - 15n + 14 = 5 , which is prime.

Case n - 2 = -1
That would be n = 1
Then 4n² - 15n + 14 = 3, which is also prime.

Case 4n - 7 = 1
That would be n = 2 and that makes other factor (n-2) zero so it's not prime

Case 4n-7 = -1
That would be n = 3/2 which is not integer, so 4n² - 15n + 14 will not be interger.

For any other n values, 4n² - 15n + 14 will be composite number since it is product of two factors.

Therefore we are left with n = 1 and n = 3; only two values of n.

5 0

2 years ago

What is the common difference or common ratio for the sequence: 27, 9, 3, 1, 1/3, ...

lawyer [7]

Answer:

It's dividing by 3 everytime, so 1/3

Step-by-step explanation:

4 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

Star Wars land encompasses an area of 14.0 acres. [1.00 acre = 4046.86m2]. If Star Wars land were made into a circle, what would

Free_Kalibri [48]

Answer:

<h3>The answer is 134.29 m</h3>

Step-by-step explanation:

First of all we need to convert 14.0 acres to m²

1.00 acre = 4046.86 m²

14.0 acres = 14 × 4046.86 = 56656.04 m²

Area of a circle = πr²

where

r is the radius

To find the radius substitute the value for the area into the above formula and solve for the radius

That's

$56656.04 = \pi {r}^{2}$

Divide both sides by π

We have

${r}^{2} = \frac{56656.04}{\pi}$

Find the square root of both sides

$r = \sqrt{ \frac{56656.04}{\pi} }$

r = 134.29139

r = 134.29 m to 2 decimal places

Hope this helps you

3 0

2 years ago