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3 years ago

Find the equation of the line that passes through the pair of points. Write your answer in slope-intercept form. (-3,0) (0,5)

Mathematics

1 answer:

Hatshy [7]3 years ago

6 0

Answer:

y= (5/3) x + 5

Step-by-step explanation:

y = mx + b

m = (5 - 0)/(0 - (-3)) = 5/3

y = (5/3)x + b

b = 0-(5/3)(-3) = 5

y = (5/3)x + 5

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You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta

vladimir1956 [14]

Answer:

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

Step-by-step explanation:

The slope of the tangent line at a point of the curve is:

$m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$m = -\frac{\cos 2\theta}{\sin \theta}$

The tangent line is horizontal when $m = 0$ . Then:

$\cos 2\theta = 0$

$2\theta = \cos^{-1}0$

$\theta = \frac{1}{2}\cdot \cos^{-1} 0$

$\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right)$ , for all $i \in \mathbb{N}_{O}$

$\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}$ , for all $i \in \mathbb{N}_{O}$

The first four solutions are:

x: $\sqrt{2}$ $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$

y: 1 -1 1 -1

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

6 0

3 years ago

What +1/16 = 3/2 answer please

Law Incorporation [45]

Answer:

1 7/16

Step-by-step explanation:

5 0

2 years ago

Given a soda can with a volume of 15 and a diameter of 2, what is the volume of a cone that fits perfectly inside the soda can?

Naddika [18.5K]

Check the picture below.

so, a cone, that fits inside the cylinder, will have the same height "h" and radius "r". Thus

$\bf \textit{volume of a cylinder}\\\\
V=\pi r^2 h\qquad \boxed{\pi r^2 h}=15
\\\\\\
\textit{volume of a cone}\\\\
V=\cfrac{\pi r^2 h}{3}\implies V=\cfrac{\boxed{15}}{3}$