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hammer [34]
3 years ago
12

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Mathematics
2 answers:
Nataly [62]3 years ago
5 0

Answer:

Step-by-step explanation:

2y - 5(y-3) = 2y -5y +15 = -3y +15

Vladimir79 [104]3 years ago
4 0

I'm guessing you want it factored since there is no solution. I got -3y+15 and its factored form would be -3(y-5).

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A polynomlal has been factored, as SHOWI Deluw.
Tom [10]

Answer:

The zeros are 6,-5,9

Step-by-step explanation:

The factored form of the polynomial is given as:

f(x) = (x - 6)(x + 5)(x - 9)

To find the zeros of this function, we set f(x)=0 and solve for x.

(x - 6)(x + 5)(x - 9) = 0

This implies that:

(x - 6) = 0 \: or \: (x + 5) = 0 \: or \: (x - 9) = 0

We solve for x to get:

x  = 6 \: or \: x  =  -  5  \: or \: x =  9

The zeros are 6,-5,9

3 0
3 years ago
What.is.the.volume....
Molodets [167]

Answer:

180

Step-by-step explanation:

V=lwh

3=10·9·6

3=180

5 0
3 years ago
A bank account earns interest at a rate of 3.5% per year ( in other words it increases in value by that percent) and starts with
LenKa [72]
Annually cumulating interest can be determined by the following formula:

W = P(1+r)^{y}

r represents the interest rate as a decimal, and P represents the starting amount of money.
8 0
3 years ago
Use an x&y to graph the function below f(x) = - (x+2)^3+1
kenny6666 [7]

Answer: f=26

Step-by-step explanation: x^2+6x^2+12x+7/x= 26

3 0
3 years ago
Evaluate the surface integral:S
rjkz [21]
Assuming S does not include the plane z=0, we can parameterize the region in spherical coordinates using

\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle

where 0\le u\le2\pi and 0\le v\le\dfrac\pi/2. We then have

x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v
(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v

Then the surface integral is equivalent to

\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du

We have

\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle
\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle
\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle
\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v

So the surface integral is equivalent to

\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du
=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv
=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw

where w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv.

=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}
=\dfrac{243}2\pi
4 0
3 years ago
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