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2 years ago

Area of square = 135 mm² Perimeter of square=

Mathematics

2 answers:

dimulka [17.4K]2 years ago

5 0

Answer:

P≈46.48

A= Area

135

Step-by-step explanation:

11Alexandr11 [23.1K]2 years ago

3 0

Answer:

About 11.62 by 11.62

A square had equal sides so you find the square root of 135 and that is rounded to 11.62.

So the perimeter would be about 46. But 46.48 exactly.

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Ashley multiplied 9 x 0.942 and got 84.78. How can you use estimation to show that Ashley's answer is wrong? Use whole numbers t

exis [7]

Answer:

Calculate the percent value:

0.942 =

0.942 × 100/100 =

(0.942 × 100)/100 =

94.2/100 =

94.2%;

In other words:

1) Multiply that number by 100.

2) Add the percent sign % to it.

Answer:

0.942 = 94.2%

Step-by-step explanation:

7 0

3 years ago

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

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

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

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

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

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

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

6 0

3 years ago

65/p • 3 when p = 5

Makovka662 [10]

Answer:

Step-by-step explanation:

65/5=13

13x3=39

8 0

3 years ago

Read 2 more answers

A cuboid measures 6cm by 3cm by 2cm Draw a net of the cuboid

Tresset [83]

Answer:

See attachment for cube

Step-by-step explanation:

Given

$Length = 6cm$

$Width = 3cm$

$Height = 2cm$

Required

The net of the cuboid

A cuboid has 6 faces.

The dimension of these faces are:

Length and Width; Length and Height;

Width and Height; Width and Height;

This means that the faces will have the following dimensions:

6cm by 3cm; 6cm by 2cm

3cm by 2cm; 3cm by 2 cm

Taking into account the above measurements and dimensions, you draw 6 faces, then join similar sides lengths.

See attachment for net of the cuboid

4 0

2 years ago

An object moves along a coordinate axis so that its position at time t is given by s(t)=t3. Estimate its instantaneous velocity

Mademuasel [1]

If the position at time t is

s(t) = (1 m/s³) t ³

then the average velocity over t = 2 s and t = 2.001 s is

v (ave) = (s (2.001 s) - s (2 s)) / (2.001 s - 2 s)

v (ave) = ((1 m/s³) (2.001 s)³ - (1 m/s³) (2 s)³) / (2.001 s - 2 s)

v (ave) ≈ (8.01201 m - 8 m) / (0.001 s)

v (ave) ≈ 12.006 m/s

6 0

2 years ago