Ask question

Ask question

All categories

3 years ago

11

The lowest elevation in Louisiana is -68 feet. The lowest elevation in California is 214 feet lower than the lowest elevation in

Louisiana. What is the lowest elevation in California?

1 answer:

Westkost [7]3 years ago

5 0

This can be expressed as x=-68-214 which is equal to -282 ft

You might be interested in

Which triangles are congruent? How do you know?

choli [55]

Use the side lengths! C and D have lengths of 3 and heights of 4.
Also use how far away they are from the axis.

Both C and D are 2 away from each axis on the end point.
This also goes with A.
However, B does not seem to be congruent in the sense of (coordinates)
It is the same length and height.

3 0

3 years ago

Read 2 more answers

Point M is the midpoint of segment AB. If AM = 50 – 3x and MB = –20 + 4x, find the value of AM

love history [14]

The answer to your question is 20

4 0

2 years ago

Read 2 more answers

44+88= __ (2+4)<br><br> 11<br> 6<br> 2<br> 22

DENIUS [597]

Dang I was just about to say 22

8 0

2 years ago

Read 2 more answers

−5x 5 y 5 (−6xy 3 ) simplify

Readme [11.4K]

Answer:

|-2| + 2

Step-by-step explanation:

-4 / 2-5<em>i</em>

cos (<em>x</em>) / 1 - sin squared (<em>x</em>)

<u>|-2| + 2</u>

8 0

1 year ago

Read 2 more answers

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in <em>c)</em>

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago