All categories

3 years ago

What is the area of the trapezoid? 80m2 96m2 100m2 120 m2

Mathematics

1 answer:

never [62]3 years ago

5 0

Answer:

100m^2

Step-by-step explanation:

area of trapezoid formula: $({b_{1} +b_{2})}{1/2h}$

b=base

base #1= 8

base #2= 12

h(height)= 10

(8+12)=20(sum of bases)

20/2=10(sum of bases over 2)

10*10=100(height * whats remaining of bases)

100m^2

hope this helps

You might be interested in

Hi I need help asap I don’t understand it can u help me for a and b

tankabanditka [31]

Answer:

Hey If you get the answer can you help me bro

Step-by-step explanation:

I am not understanding that either

6 0

2 years ago

5. State the missing reasons in this proof.

hammer [34]

Answer:

a.) Vertical Angles

b.) Given (Parallel)

c.) Substitution

Step-by-step explanation:

3 0

2 years ago

Can someone help me real quick?

Tpy6a [65]

3,400×(1+0.04)^(7)=4,474.17

4 0

3 years ago

Consider two people where one person is 25% taller than the other but proportioned in exactly the same way. (That is, the tall

egoroff_w [7]

Answer:

46.25

Step-by-step explanation:

If they are proportioned exactly the same, then everything on the taller person is 25% larger

37*1.25 (for them being 125% the size of the smaller person)

46.25

4 0

2 years ago

Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = 2yi + xzj + (x

Lady_Fox [76]

By Stokes' theorem, the line integral of $\vec F$ over $C$ is given by the surface integral of the curl of $\vec F$ over $S$ , where $S$ is the region of intersection of the plane $z=y+6$ and the cylinder $x^2+y^2=1$ with $S$ having positive/upward orientation.

Parameterize $S$ by

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

with $0\le u\le1$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath+u\,\vec k$

The curl of $\vec F$ is

$\nabla\times\vec F(x,y,z)=(1-x)\,\vec\imath-\vec\jmath+(z-2)\,\vec k$

Then the line integral is equivalent to

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^1\bigg((1-u\cos v)\,\vec\imath-\vec\jmath+(u\sin v+4)\,\vec k\bigg)\cdot\bigg(-u\,\vec\jmath+u\,\vec k\bigg)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^1(5u+u^2\sin v)\,\mathrm du\,\mathrm dv=\boxed{5\pi}$

4 0

2 years ago