All categories

3 years ago

What is the area of a triangle with a base of 10 and height of 4?

Mathematics

1 answer:

PilotLPTM [1.2K]3 years ago

6 0

The area of a triangle with a base of 10 and a height of 4 would end up being 20

You might be interested in

Consider th following formula used to find a cone v= 3.14r^2 h/3

lys-0071 [83]

Here's the volume formula (and the area formula) typed a little neater than that.
(I'm guessing you forgot to type in your question?)

7 0

3 years ago

Solve for x.<br> X + 10<br> 2x - 50<br><br> x = [?]

Vladimir [108]

Answer:

x +10 = 2x-50

x-2x = -50-10

-x. = -60

-x/-1 = -60/-1

x = 60°

6 0

3 years ago

A number line is divided into fifths. How many equal segments are there between the point at zero and the point 4/5?

ella [17]

Answer:

Step-by-step explanation:

A little confused by the question but is it just

1/5 2/5 3/5 4/5

1 2 3 4

so theres 3 in between

7 0

3 years ago

Question 5 of 10

Lady_Fox [76]

Answer:

______

x = − 0. 142857

Step-by-step explanation:

7 0

2 years ago

Evaluate the surface integral ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S f (x

kykrilka [37]

Parameterize $S$ by

$\vec s(u,v)=6\cos u\sin v\,\vec\imath+6\sin u\sin v\,\vec\jmath+6\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take a normal vector to $S$ ,

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=36\cos u\sin^2v\,\vec\imath+36\sin u\sin^2v\,\vec\jmath+36\cos v\sin v\,\vec k$

which has norm

$\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|=36\sin v$

Then the integral of $f(x,y,z)=x^2+y^2$ over $S$ is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\iint_S\left((6\cos u\sin v)^2+(6\sin u\sin v)^2\right)\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv$

$=\displaystyle36^2\int_0^{\pi/2}\int_0^{2\pi}\sin^3v\,\mathrm du\,\mathrm dv=\boxed{1728\pi}$

6 0

3 years ago