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2 years ago

Distribute this expression: ab(5c+9)

Mathematics

2 answers:

UNO [17]2 years ago

7 0

Answer:

9b+5cb

Step-by-step explanation:

irina [24]2 years ago

5 0

9b+5cb would be your answer

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Catron evaluates the expression (negative 9) (2 and two-fifths) using the steps below.

Vlad [161]

Answer:

Catron's error is

"She did not follow order of operations"

Step-by-step explanation:

Catron evaluates the expression (negative 9) (2 and two-fifths)

That expression can be written as below

$(-9)(2\frac{2}{5})$

Catron's error is

"She did not follow order of operations"

The corrected steps are

Step1: Given expression is $(-9)(2\frac{2}{5})$

Step2: Convert mixed fraction into improper fraction

$(-9)(2\frac{2}{5})=(-9)(\frac{12}{5})$

Step3: Multiplying the terms

$(-9)(2\frac{2}{5})=-7\frac{-108}{5}$

Therefore solution $(-9)(2\frac{2}{5})=-7\frac{-108}{5}$

6 0

3 years ago

Read 2 more answers

Can someone take 3 min for there time and help me with this please??

seropon [69]

Since the left side of the equation says f(5), then plug in 5 for x.

So, it is:

-4|5| + 3
= -20 + 3
= 23

4 0

3 years ago

Need help number 8 it’s decreasing or increasing

andrew-mc [135]

Increasing I think dryyhiiouiiiiuyutff

5 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

I ONLY HAVE 2 MINS PLS HELP

stellarik [79]

Answer:

Forgive me if I am wrong, but I beleive the answer is F

Step-by-step explanation:

7 0

3 years ago