Ask question

Ask question

All categories

3 years ago

10

What does a decreasing slope means in terms of distance?

1 answer:

Sergio [31]3 years ago

7 0

Answer:

Step-by-step explanation:

In terms of distance, a decreasing slope means a slowing speed, which leads to a shorter distance.

You might be interested in

(please Explain each step)<br>the question is on the image.

Fiesta28 [93]

d= distance travelled = 6 km

P= 5d²+270

= 5(6)²+270

=5 ×36 +270

= 180+ 270

= 450

Taxi fare = 450 pence

7 0

3 years ago

Tyrone is constructing a box in the shape of a rectangular prism to store his baseball cards. it has a length of 10 centimeters

kupik [55]

The formula for this, as you stated, is Volume = Length * Height * Width.

Therefore, the Volume is 10*7*8, or 560

Therefore, the volume of the box is 560 cm^3

Hope this helped!! :D

6 0

3 years ago

Katie's swim team is having an end-of-season pizza party. When the party started, there was 50 slices of pizza. Now there are 20

kotykmax [81]

Answer:

30alice of pizza they eat

5 0

3 years ago

if a pair of shoes cost Rs 207.50 and a pair of socks Rs 42.50 how many set of shoes and shocks can be brought with Rs 3000

Bezzdna [24]

You can buy 12 sets of shoes and socks

7 0

2 years ago

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1x2+y2≤1. the plate, including t

vredina [299]

You're looking for the extreme values of $x^2+3y^2+13x$ subject to the constraint $x^2+y^2\le1$ .

The target function has partial derivatives (set equal to 0)

$\dfrac{\partial(x^2+3y^2+13x)}{\partial x}=2x+13=0\implies x=-\dfrac{13}2$

$\dfrac{\partial(x^2+3y^2+13x)}{\partial y}=6y=0\implies y=0$

so there is only one critical point at $\left(-\dfrac{13}2,0\right)$ . But this point does not fall in the region $x^2+y^2\le1$ . There are no extreme values in the region of interest, so we check the boundary.

Parameterize the boundary of $x^2+y^2\le1$ by

$x=\cos u$

$y=\sin u$

with $0\le u$ . Then $t(x,y)$ can be considered a function of $u$ alone:

$t(x,y)=t(\cos u,\sin u)=T(u)$

$T(u)=\cos^2u+3\sin^2u+13\cos u$

$T(u)=3+13\cos u-2\cos^2u$

$T(u)$ has critical points where $T'(u)=0$ :

$T'(u)=-13\sin u+4\sin u\cos u=\sin u(4\cos u-13)=0$

$(1)\quad\sin u=0\implies u=0,u=\pi$

$(2)\quad4\cos u-13=0\implies\cos u=\dfrac{13}4$

but $|\cos u|\le1$ for all $u$ , so this case yields nothing important.

At these critical points, we have temperatures of

$T(0)=14$

$T(\pi)=-12$

so the plate is hottest at (1, 0) with a temperature of 14 (degrees?) and coldest at (-1, 0) with a temp of -12.

4 0

3 years ago