All categories

4 years ago

What is the equation of the line that has a slope of -4 and goes through (-1,6)?

Mathematics

1 answer:

OLga [1]4 years ago

3 0

Answer:

y = - 4x + 2

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = - 4, thus

y = - 4x + c ← is the partial equation

To find c substitute (- 1, 6) into the partial equation

6 = 4 + c ⇒ c = 6 - 4 = 2

y = - 4x + 2 ← equation of line

You might be interested in

Given that a x 70 = b work out the value of 3b/a Your final line should say, 3b/a = ...

Leto [7]

Answer:

$\frac{3b}{a} = 210$

Step-by-step explanation:

Explanation:-

Given that a× 70 = b

Dividing 'a' into both sides, we get

⇒ $\frac{b}{a} = 70$

we have to find the value $\frac{3b}{a}$

$\frac{3b}{a} = 3 X \frac{b}{a}$

= 3× 70

= 210

3 0

3 years ago

What is the value of sin 15

Lena [83]

Answer:

sin(15 radians) is 0.65028784015

4 0

3 years ago

Read 2 more answers

A flat circular plate has the shape of the region x2 + y2≤1. The plate, including the boundary where x2 + y2 = 1, is heated such

olganol [36]

$T(x,y)=x^2+2y^2-x$
$\implies\nabla T(x,y)=(T_x,T_y)=(2x-1,4y)$

Setting both partial derivatives to 0 gives a single critical point at $(x,y)=\left(\dfrac12,0\right)$ , which does fall inside the unit disk.

At this point, the value of the derivative of the Hessian matrix is

$|H|=\begin{vmatrix}T_{xx}&T_{xy}\\T_{yx}&T_{yy}\end{vmatrix}=\begin{vmatrix}2&0\\0&4\end{vmatrix}=8>0$

while the value of the second-order partial derivative with respect to $x$ is

$T_{xx}\bigg|_{(x,y)=(1/2,0)}=2>0$

This means the critical point is the site of a local minimum, so this is the coldest point on the plate with a temperature of $T\left(\dfrac12,0\right)=-\dfrac14$ .

The hottest point on the plate must then be found on the boundary. Let $x=\cos\theta$ and $y=\sin\theta$ , so that

$T(x,y)=T(\theta)=\cos^2\theta+2\sin^2\theta-\cos\theta$
$T(\theta)=\dfrac32-\cos\theta-\dfrac12\cos2\theta$

Then the boundary of the plate (the circle $x^2+y^2=1$ ) is a function of a single variable $\theta$ considered over $\theta\in[0,2\pi)$ . Differentiating once gives

$T'(\theta)=\sin\theta+\sin2\theta=0$
$\implies\theta=0,\theta=\dfrac{2\pi}3,\theta=\pi,\theta=\dfrac{4\pi}3$

You'll find that $T(\theta)$ attains three extrema on the interval $(0,2\pi)$ , with relative maxima at $\theta=\dfrac{2\pi}3$ and $\theta=\dfrac{4\pi}3$ and a relative minimum at $\theta=\pi$ (and $\theta=0$ , if you want to include that).

We already found our minimum on the inside of our plate - which you can verify to have a lower temperature than at the points given by $T(\theta)$ - and we find two maxima at $\theta=\dfrac{2\pi}3$ and $\theta=\dfrac{4\pi}3$ , each giving a maximum temperature of $T=\dfrac94$ .

Converting back to Cartesian coordinates, these points correspond to the points $\left(-\dfrac12,\pm\dfrac{\sqrt3}2\right)$ .