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2 years ago

-2 1/3+(1 3/4 I need help for a test so if you guys can help me I'll be happy :D

Mathematics

1 answer:

kotegsom [21]2 years ago

6 0

The answer is -7/12! Please mark me brainest! :)

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10ft 19 ft how do you do this

tino4ka555 [31]

If we're talking about a box with dimensions 10 ft, 19 ft and 20 ft, it's probable that the instructions say to calculate the volume. That would be

V = (10 ft)(19 ft)(20 ft) = 3800 cubic feet.

8 0

2 years ago

Can someone translate this into a verbal expression 3(4j + 4 + j)?

dalvyx [7]

3(5j+4)
15j+12

final answer 15j+12

5 0

2 years ago

I want to install carpet in my living room, which is 200 inches long and 50 ft wide. The carpet is sold for 5000 naira per yard2

inn [45]

Answer:

462962.96 naira

Step-by-step explanation:

This problem is complicated by the use of two different units. I chose to use inches for the dimensions of the carpet.

200 in long

50 ft wide = 50 x 12 = 600 in.

Area = 200 x 600 = 120000 in^2

A square yard is a square 36 in. x 36 in. = 1296 in^2

Divide 120000 by 1296 to find out how many square yards of carpet you'll need.

120000 / 1296 = 92.59

Multiply by 5000 naira/yd^2 = 462962.96

7 0

2 years ago

The surface area of this triangular pyramid, whose base and lateral faces are congruent equilateral triangles, is __ square inch

DedPeter [7]

Answer:

The surface area is equal to $390\ in^{2}$

Step-by-step explanation:

The surface area of the triangular pyramid is equal to the area of its four triangular faces

In this problem

$SA=4[\frac{1}{2}(b)(h)]$

we have

$b=15\ in$

$h=13\ in$

substitute the values

$SA=4[\frac{1}{2}(15)(13)]=390\ in^{2}$

4 0

3 years ago

Read 2 more answers

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