All categories

2 years ago

This graph shows the number of miles traveled by a train over a 5 hour trip.

Mathematics

2 answers:

IrinaK [193]2 years ago

8 0

D. It would be 40 mph because if you take 180 divided by 40 it equals 5

12345 [234]2 years ago

6 0

Answer:

35 mph

Step-by-step explanation:

if you look at the 1 hour mark its at 35 and the 2 hour is at 70

70/2=35

I hope this helps!!!

You might be interested in

What the best subject

AURORKA [14]

Mathmatic is what most people ask questions about, but the best for this site is math or English. Hope this helped.

3 0

2 years ago

Read 2 more answers

Find the equation of a line that passes through the point (4,15) and is perpendicular to the equation below.

mr Goodwill [35]

The answer is D because to find the slope of a line that is perpendicular to another, you have to find the opposite reciprocal. so here the slope would be -1/4. now, take that slope and put it into an equation. like the following:
$y = \frac{ - 1}{4} x + b$
now to find b you must insert the points into the equation (plug in 4 for x, and 15 for y).
$15 = \frac{ - 1}{4} (4) + b$
b then equals 16. so, the answer is D.

8 0

2 years ago

Find the slope of the line graphed below picture is attached

777dan777 [17]

Answer: The answer is D.

3 0

1 year ago

Read 2 more answers

Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin

saw5 [17]

Treat $\mathcal C$ as the boundary of the region $\mathcal S$ , where $\mathcal S$ is the part of the surface $z=2xy$ bounded by $\mathcal C$ . We write

$\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r$

with $\mathbf f=(y+7\sin x,z^2+9\cos y,x^3)$ .

By Stoke's theorem, the line integral is equivalent to the surface integral over $\mathcal S$ of the curl of $\mathbf f$ . We have

$\nabla\times\mathbf f=(-2z,-3x^2,-1)$

so the line integral is equivalent to

$\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$
$=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

where $\mathbf s(u,v)$ is a vector-valued function that parameterizes $\mathcal S$ . In this case, we can take

$\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)$

with $0\le u\le1$ and $0\le v\le2\pi$ . Then

$\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$

and the integral becomes

$\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$
$=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi$

4 0

2 years ago

Use substitution to solve the following system of equations. What is the value of y? {3x+2y=12{5x−y=7 A) y = -3B) y = 3C) y = -2

Vlad [161]

Answer

B) y = 3

Step-by-step explanation

Given the system of equations:

$\begin{gathered} 3x+2y=12\text{ \lparen eq. 1\rparen} \\ 5x-y=7\text{ \lparen eq. 2\rparen} \end{gathered}$

Isolating x from equation 1:

$\begin{gathered} 3x+2y-2y=12-2y \\ 3x=12-2y \\ \frac{3x}{3}=\frac{12-2y}{3} \\ x=\frac{12}{3}-\frac{2}{3}y \\ x=4-\frac{2}{3}y\text{ \lparen eq. 3\rparen} \end{gathered}$

Substituting equation 3 into equation 2 and solving for y:

$\begin{gathered} 5(4-\frac{2}{3}y)-y=7 \\ 5\cdot4-5\cdot\frac{2}{3}y-y=7 \\ 20-\frac{10}{3}y-y=7 \\ 20-\frac{13}{3}y=7 \\ 20-\frac{13}{3}y-20=7-20 \\ -\frac{13}{3}y=-13 \\ (-\frac{3}{13})\cdot-\frac{13}{3}y=(-\frac{3}{13})\cdot-13 \\ y=3 \end{gathered}$

6 0

7 months ago