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3 years ago

7

A jigsaw puzzle contains 50 pieces. If joining any two pieces or groups is considered one move, what is the least amount of move

s to join all 50 pieces?

1 answer:

Nina [5.8K]3 years ago

5 0

So, you see that when 2 pieces is considered a group, that means you have to divide. So, you divide 50/2 and it gives you 25.

Hope this helps :)

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Solve the problem, calculate the line integral of f along h

Over [174]

The curve $\mathcal H$ is parameterized by

$\begin{cases}X(t)=R\cos t\\Y(t)=R\sin t\\Z(t)=Pt\end{cases}$

so in the line integral, we have

$\displaystyle\int_{\mathcal H}f(x,y,z)\,\mathrm ds=\int_{t=0}^{t=2\pi}f(X(t),Y(t),Z(t))\sqrt{\left(\frac{\mathrm dX}{\mathrm dt}\right)^2+\left(\frac{\mathrm dY}{\mathrm dt}\right)^2+\left(\frac{\mathrm dZ}{\mathrm dt}\right)^2}\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}Y(t)^2\sqrt{(-R\sin t)^2+(R\cos t)^2+P^2}\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}R^2\sin^2t\sqrt{R^2+P^2}\,\mathrm dt$
$=\displaystyle\frac{R^2\sqrt{R^2+P^2}}2\int_0^{2\pi}(1-\cos2t)\,\mathrm dt$
$=\pi R^2\sqrt{R^2+P^2}$

You are mistaken in thinking that the gradient theorem applies here. Recall that for a scalar function $f:\mathbb R^n\to\mathbb R$ , we have gradient $\nabla f:\mathbb R^n\to\mathbb R^n$ . The theorem itself then says that the line integral of $\nabla f(x,y,z)=\mathbf f(x,y,z)$ along a curve $C$ parameterized by $\mathbf r(t)$ , where $a\le t\le b$ , is given by

$\displaystyle\int_C\mathbf f(x,y,z)\,\mathrm d\mathbf r=f(\mathbf r(b))-f(\mathbf r(a))$

Specifically, in order for this theorem to even be considered in the first place, we would need to be integrating with respect to a vector field.

But this isn't the case: we're integrating $f(x,y,z)=y^2$ , a scalar function.

7 0

3 years ago

Subtract -6x^4-3x^2y^2+5y^4−6x

Gnoma [55]

Answer:

$8x^4-5x^2y^2-6y^4+6x$

Step-by-step explanation:

We are required to subtract $-6x^4-3x^2y^2+5y^4-6x \:from \: 2x^4-8x^2y^2-y^4$ .

$[2x^4-8x^2y^2-y^4]-[-6x^4-3x^2y^2+5y^4-6x]$

<u>Step 1:</u> Open the bracket, Take note of the signs

$[2x^4-8x^2y^2-y^4]-[-6x^4-3x^2y^2+5y^4-6x]\\2x^4-8x^2y^2-y^4+6x^4+3x^2y^2-5y^4+6x$

<u>Step 2:</u> Collect like terms

$2x^4+6x^4-8x^2y^2+3x^2y^2-y^4-5y^4+6x$

<u>Step 3:</u> Simplify

$8x^4-5x^2y^2-6y^4+6x$

5 0

3 years ago

Melinda and Marcus each have a job. Melinda earns $12 an hour. She also receives a $75 weekly bonus.

Olenka [21]

According to given information,
Equation of Melinda Earning = 75 + 12x
Equation of Marcus Earning = 15x
Where, x = Number of hours they worked.

Now, we need to calculate when they will have same amount of money, so for that instance they shall equal to each other.
75 + 12x = 15x
15x - 12x = 75
3x = 75
x = 75/3
x = 25

After 25 hours, they will have the money = 15(25) = $375

So, your final answer is $375

Hope this helps!

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3 years ago

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Solve for X<br> 121 +7< -11<br> OR 5x- 8 > 40

Trava [24]

Answer:

x > 48/5 the first one is false

3 0

3 years ago

If gumballs cost 150 a bag ,and there 15!gumballs in each bag much do you need to charge for each gumball in order to make money

pshichka [43]

More than 10 a gumball.

150 divided by 15 is 10, which would be the price you bought for. If you want to make more money, you must at least have a surge pricing of 10 unit prices per gumball.

7 0

3 years ago

Read 2 more answers