Help!!! What is the perimeter of the triangle?

X² y² + y2 – 4x + 20y +55 =0<br> Center:<br> Radius:

netineya [11]

Answer:

ya

Step-by-step explanation:

22

4 0

2 years ago

A stone is thrown upward from ground level. With what minimum speed should the stone be thrown so as to reach a height of 36 fee

rusak2 [61]

V^2 = u^2 - 2gh; where v is the final speed, u is the initial speed, g = 9.8 m/s = 0.54 ft/min is the accerelation due to gravity, h = height.
At the maximum height, the final speed is 0.
u^2 = 2gh = 2 x 0.54 x 36 = 38.58
u = sqrt(38.58) = 6.21 ft/min

Therefore, the stne should be thrown with 6.21 ft/min

4 0

2 years ago

Using the data shown on the graph, which statements are correct

Tomtit [17]

Answer:

B, C, and E

Step-by-step explanation:

Let's consider each option given the graph shown above:

A. The ratio of x/y can be found using any point on the line, say, (5, 6). Thus, ratio of x/y = 5/6, not 6/5.

Option A is not correct.

B. If you check for y/x for any given point along the line, we would have the same result. So the ratio of y/x is consistent. Option B is correct.

C. The graph has a line that runs through the point of origin, so therefore, it represents a proportional relationship.

Option C is correct

D. If option C is correct, then option D is not correct.

E. If option C is correct, then option E is correct.

8 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

Please helpppp!!! Will mark brainlyist

Snowcat [4.5K]

Answer:

x = 35.5

Step-by-step explanation:

cos 65° = 15/x

cos 65° · x = 15

x = 15/cos 65°

x = 35.5

4 0

2 years ago