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9966 [12]
3 years ago
8

Can we consider 5√×as monomial? ​

Mathematics
2 answers:
antoniya [11.8K]3 years ago
8 0

Answer:

A monomial is an expression in algebra that contains one term, like 3xy. Monomials include: numbers, whole numbers and variables that are multiplied together, and variables that are multiplied together. Any number, all by itself, is a monomial, like 5 or 2,700. A monomial can also be a variable, like m or b. It can also be a combination of these, like 98b or 7rxyz.

PilotLPTM [1.2K]3 years ago
5 0

Answer:

A monomial is an expression in algebra that contains one term, like 3xy. Monomials include: numbers, whole numbers and variables that are multiplied together, and variables that are multiplied together. A polynomial is a sum of monomials where each monomial is called a term.

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For 32 = 5X + 2, what is the first step in solving for X?
nikitadnepr [17]

Answer:

-2 from both sides

Step-by-step explanation:

32=5x+2

-2        -2

________

30=5x

__=__

5      5

6=x

4 0
2 years ago
True or false? f (x) = 2 · (1/5)^x represents an exponential function growth.
nekit [7.7K]

Hi there!

\large\boxed{\text{False.}}

f(x) = 2(1/5)ˣ

Recall the form of an exponential function:

f(x) = a(b)ˣ where:

a = initial value

b = rate of decay/growth

If b > 1, the function is undergoing exponential growth. If b < 1, then the function is undergoing exponential decay.

1/5 < 1, so the function is undergoing exponential decay, not growth.

7 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
Courtney had $55 to spend at the fair. She paid $22.50 for rides and $14.25 for lunch and snacks. Courtney says she has $18.25 l
IrinaVladis [17]
She has $55. She spent $22.50 on rides so now she has $32.50 . Then she spends $14.25 on lunch and snack so now she has $18.25.


A. would be wrong because Courtney IS correct, she has $18.25 left.

C. is also wrong because, again, Courtney IS correct.

So the only option left is B and it explains why Courtney is correct.

Option B is the correct answer.

Hope this helps.

4 0
3 years ago
" A projectile is launched in the air according to the equation h(t)= -16t^2 + 125t + 5. At what time does the projectile land?"
Vlad1618 [11]
"according to the equation is a sloppy description, and announces that whoever said it or wrote it doesn't really have a clue to what the equation means or what it's good for.

h (t) is the HEIGHT of the projectile above the ground at any time 't'. When the projectile hits the ground, h (t) is zero. Write that ! Then you have a quadratic equation that you can easily solve for 't'.
8 0
3 years ago
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