If a=3 and b=4 then b exponet of a

5).
and
6).
The volume of a sphere is

   (4/3) (pi) (radius)³ .

In #5, the 'pi' is already there next to the answer window.
You just have to come up with the (4/3)(radius³).

Remember that the radius = 1/2 of the diameter.

7). The volume of a cylinder is

   (pi) (radius²) (height) .

Divide the juice in the container by the volume of one can,
to get the number of cans he can fill.

8). The volume of a cone is

   (1/3) (pi) (radius of the round bottom)² (height) .

He starts with a small cone, he then adds clay to it to make it higher.
The question is: How much clay does he ADD to the short one,
to make the bigger one ?

Use the formula to find the volume of the short one.
Use the formula again to find the volume of the bigger one.
Then SUBTRACT the smaller volume from the bigger volume.
THAT's how much clay he has to ADD.

Notice that the new built-up cone has the same radius
but more height than the first cone.
_______________________________________

Don't worry if you don't understand this.

The answer will be this number:

(1/3) (pi) (radius²) (height of the big one minus height of the small one).

6 0

3 years ago

3d-4 when d=1.2<br><br><br> Please help its homework

AysviL [449]

-0.4 because it is 3.6 - 4

5 0

3 years ago

Read 2 more answers

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

7 0

2 years ago