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2 years ago

How do you explain to estimate 368+231 in two different ways

1 answer:

7 0

You can use the break apart strategy which would look like:
368+231= 300+60+8 + 200+30+1
Or you could just plain and simple add 368+231

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Sina played basketball on a rectangular court that was 74 feet by 42 feet. after the game, she walked across the court diagonall

Naddika [18.5K]

the distance Sina walked was 7,240 feet

6 0

1 year ago

Read 2 more answers

Select the statement that is true. Group of answer choices There are integers s and t such that 2=102⋅s+72⋅t . here are integers

LekaFEV [45]

Answer:

Step-by-step explanation: true

3 0

2 years ago

The weight, w, of a spring in pounds is given by 0.9 times the square root of the energy, E, stored by the spring in joules. If

frutty [35]

For this case we have the following equation:
$w = 0.9* \sqrt{E}$
Where,
w: The weight of a spring in pounds
E: the energy stored by the spring in joules.
Substituting values we have:
$w = 0.9* \sqrt{12}$
Making the corresponding calculation:
$w=3.12$
Answer:
the approximate weight of the spring in pounds is:
$w=3.12$

4 0

2 years ago

Read 2 more answers

The diagram shows some land in the shape of a quadrilateral ABCD. AB = 6 km, AD = 5 km and CD = 12 km Angle BAC = 30 The land is

Goshia [24]

Answer:

£495 million

Step-by-step explanation:

To find out the total cost of the land, we need to first calculate the area of the land.

Step 1: Find area of right angled triangle ADC

AD = 5 km,

DC = 12 km

Area of the right triangle = ½*a*b

a = 5km

b = 12km

Area = ½*5*12

= 5*6

Area of ADC = 30 km²

Step 2: Find the area of triangle ABC

First, let's find the length of AC using Pythagorean theorem

AC² = AD² + DC²

AC² = 5² + 12² = 25 + 144

AC = √169

AC = 13km

Area of ∆ABC = ½*AB*AC*sin(30°)

= ½*6*13*0.5

= 3*13*0.5

Area of ∆ABC = 19.5 km²

Total area of the land = area of ∆ADC + ∆ABC = 30 + 19.5 = 49.5 km²

Step 3: calculate how much the land costs

If the land costs £10 million per km²,

Cost of 49.5 km² = 49.5 × 10 = £495 million

4 0

2 years ago

Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 −

Korolek [52]

The vector field

$\vec F(x,y,z)=x^2\sin z\,\vec\imath+y^2\,\vec\jmath+xy\,\vec k$

has curl

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

Parameterize $S$ by

$\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k$

where

$\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}$

with $0\le u\le3$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

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

Then by Stokes' theorem we have

$\displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^3u^3(2u\cos^3v\sin(u^2-9)+\cos^3v\sin v+2u\sin^3v+\cos v\sin^3v)\,\mathrm du\,\mathrm dv$

which has a value of 0, since each component integral is 0:

$\displaystyle\int_0^{2\pi}\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin v\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\cos v\sin^3v\,\mathrm dv=0$

4 0

2 years ago