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3 years ago

How do I solve 0.6(-13k-12)

Mathematics

2 answers:

Anit [1.1K]3 years ago

7 0

Answer:

Explanation:

0.6 (-13k - 12)

<em>Apply the distributive property</em>

0.6 (-13k) - 0.6 · 12

<em>Apply minus - plus rules</em>

-13 · 0.6k - 12 - 0.6

<em>Simplify</em>

-7.8k - 7.2

Anton [14]3 years ago

4 0

Answer:

-7.8k - 7.2

Step-by-step explanation:

Distribute 0.6 to the terms in the parentheses:

0.6(-13k - 12)

-7.8k - 7.2

So, the simplified expression is -7.8k - 7.2

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The length of the rectangular basketball court is 44 feet more than the width. If the perimeter of the basketball court is 288 f

Temka [501]

Let's first say that L=W+44

and then remember that perimeter is P=2L+2W

replace the L with W+44

we then get P=2(W+44)+2W, now I'll solve it

P=2W+88+2W
P=4W+88

substitute 288 for P

288=4W+88
200=4W
50=W

so now we now how wide the court is. add 44 to find the length which gives you L=94

as always plug the numbers back into your perimeter equation to ensure L and W are correct

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3 years ago

2(5m^4n^-1)^3 simplify show work please

dangina [55]

Answer:

$\large\boxed{2(5m^4n^{-1})^3=250m^{12}n^{-3}=\dfrac{250m^{12}}{n^3}}$

Step-by-step explanation:

$\text{Use}\\\\(ab)^n=a^nb^n\\\\(a^n)^m=a^{nm}\\\\a^{-n}=\dfrac{1}{a^n}\\-------------------\\\\2(5m^4n^{-1})^3=2(5)^3(m^4)^3(n^{-1})^3=2(125)m^{3\cdot4}n^{-1\cdot3}=250m^{12}n^{-3}=\dfrac{250m^{12}}{n^3}$

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3 years ago

Question 2 Determine if (x-3) is a factor to the polynomial P(x) = 2x³ +5x²-28x - 15. (3 marks)

tangare [24]

Answer:

(x - 3) is a factor of P(x)

Step-by-step explanation:

if (x - 3) is a factor of P(x) then P(3) = 0

P(3) = 2(3)³ + 5(3)² - 28(3) - 15

= 2(27) + 5(9) - 84 - 15

= 54 + 45 - 99

= 99 - 99

= 0

since P(3) = 0 then (x - 3) is a factor of P(x)

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2 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

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

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

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

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$