Ask question

Ask question

All categories

3 years ago

7

Talia took the bus from her home to the bank and then walked back to her home along the same route. The round trip took 0.9 hour

s total. The bus traveled at an average speed of 40 km/h and she walked at an average speed of 5 km/h. Use the table to complete these statements. The rate of Trip 2 is km/h. The time of Trip 1 is hours.

1 answer:

WARRIOR [948]3 years ago

4 0

It is 5 then 0.9-x on E2020

You might be interested in

Find dy/dx if y=(4x^2-5x)^8

musickatia [10]

dy/dx = 8(8x - 5)(4x² - 5x )^ 7

differentiate using the ' chain rule '

dy/dx = 8(4x² - 5x)^ 7 × d/dx (4x² - 5x)

= 8(4x² - 5x)^ 7 × (8x - 5)

= 8(8x - 5)(4x² - 5x) $^{7}$

6 0

2 years ago

Read 2 more answers

debt payments of $800 due now and $1400 due in 5 months are to be repaid by a payment of $1000 in 3 months and a final payment i

loris [4]

Answer:

$1,304.70

Step-by-step explanation:

If interest 6% annually, monthly is 0.5%.

The debt in 5 months will be 800 plus compounded interest for 5 months plus new due debt

$800(1.005)^5+1400=2220.201$

In 3 more months the debt will be 2220.201 plus compounded interest for 3 months minus payment

$2220.201(1.005)^3-1000=1253.67$

After 8 months the debt would be 1253.67 plus compounded interest for 8 months

$2220.201(1.005)^3=1304.70$

Then the size of the final payment would be $1,304.70

5 0

3 years ago

Find the Area of ze Trapezoid

Elina [12.6K]

A=a+b/2·h=9+4/2·3.5=22.75

Hoped I helped!

3 0

3 years ago

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt$

Complete the square in the quadratic term of the integrand: $t^2-2t+2=(t-1)^2+1=(1-t)^2+1$ , then in the integral we substitute $u=1-t$ :

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt$

$\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du$

Make another substitution of $v=u^2$ :

$\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv$

Integrate by parts, taking

$r=v+1\implies\mathrm dr=\mathrm dv$

$\mathrm ds=e^v\,\mathrm dv\implies s=e^v$

$\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv$

$\displaystyle=-(2e-1)+(e-1)=-e$

So, we have by the fundamental theorem of calculus that

$\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)$

$\implies-e=f(1,1,1)-2$

$\implies f(1,1,1)=2-e$

3 0

3 years ago

Help me with this please I will give you brainlist!

Volgvan

Answer:

it might be D

Step-by-step explanation:

im pretty sure

4 0

2 years ago