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3 years ago

Can someone answer my last question please?

Mathematics

1 answer:

max2010maxim [7]3 years ago

6 0

What was your last question, I might be able to help.

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On a number line what is the distance between -12 and 20 ??

KIM [24]

Answer: Step-by-step explanation:

In order to answer this question you will need to find 2 distances and then add them to each other.

Begin at -29. How far will you have to move to the right 29 in order to get to 0?

Now how far will you have to more to get from 0 to 100?

Draw an open number line to help you.

_____________________________________________________

-29 0 100

Your first distance is 29. Your second distance is 100. The total distance is 129.

3 0

3 years ago

Sketch the graph of each line

adoni [48]

Answer:

(0,5) (1,0)

Step-by-step explanation:

Place your first dot on the 5 (0,5) of the y-axis (The one that is vertical) from there count down the line one value at a time until you have counted DOWN 5 times. Go to the right once and place another point there (1,0). Then draw a straight line that connects those two points.

6 0

2 years ago

Pls solve it and show your work :((

lisabon 2012 [21]

Answer:

See below

Step-by-step explanation:

To graph the line we need two points, one point is the y-intercept, the second point to be calculated.

y = -3x + 6

The y-intercept is (0, 6)

Let the domain be x = 5 for the second point, then:

y = -3*5 + 6 = -15 + 6 = -9
The point is (5, -9)

Connect these points to get the graph

y = 4/5x - 3

The y-intercept is (0, -3)

Let the domain be x = 5 for the second point, then:

y = 4/5*5 - 3 = 4 - 3 = 1
The point is (5, 1)

Connect these two points to get the graph

5 0

2 years ago

The width of a rectangular room is 60 percent of its length. the length of the room is 10 feet. what is the area of the room

blsea [12.9K]

The width is 6 feet because 10•0.6=6

4 0

3 years ago

Read 2 more answers

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago