How far is a million inches in miles

In triangle ABC, the measure of angle B is 50 degrees.

beks73 [17]

Sup do you have the photo of your hw?

8 0

2 years ago

Identical sweaters are on sale in two different stores. The sale price in store A is 30% off the regular price of $25. The sale

bekas [8.4K]

30% off 25: multiply 25 by .30 to find 30%: 7.5. Now subtract that from the sales price: 25-7.5=17.5
Do the same for store B: .40•30=12. 30-12=18.
So store A is better by $.50 :) hope this helps :)

7 0

2 years ago

Read 2 more answers

Solve for x. x= In e x=

IgorLugansk [536]

Good morning ☕️

Answer:

x = 1

Step-by-step explanation:

since the function ln is the inverse of the function exponential then :

$ln(e )=lne^{1} = 1$

:)

3 0

2 years ago

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

Let f(x)=x^3-x-1

alexira [117]

$f(x) = x^3 - x - 1$

To find the gradient of the tangent, we must first differentiate the function.

$f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1$

The gradient at x = 0 is given by evaluating f'(0).

$f'(0) = 3(0)^2 - 1 = -1$

The derivative of the function at this point is negative, which tells us the function is decreasing at that point.

The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so

$y = -x + c$

Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).

$f(0) = (0)^3 - (0) - 1 = -1$

So the point (0, -1) lies on the tangent. Substituting into the tangent equation:

$y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}$

6 0

3 years ago