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KonstantinChe [14]
3 years ago
5

Classify this by degree and by number of terms?

Mathematics
1 answer:
sveta [45]3 years ago
3 0
The answer is quintic binomial.

It is quintic because its highest power is 5. Quartic would mean its highest power is 4.

It is a binomial because the two terms are -4x^5 and -6x^3. Terms are separated by + or - signs.
You might be interested in
Write the equation of the graph which is the reflection of y=−|x−1|−1 about the origin
Norma-Jean [14]

Answer:

f(x) = |x + 1| + 1

Step-by-step explanation:

A reflection over the y-axis followed by a reflection over x-axis is equivalent to a reflection about origin.

f(x) = -|x-1| -1

reflection over the x-axis:

-f(x) = (-1)*(-|x-1| -1) = |x-1| +1

reflection over the y-axis of the previous function:

-f(-x) = |-x-1| +1 = |x + 1| + 1

which is the reflection of f(x) about the origin

6 0
3 years ago
Outline the steps for taking cross sections of a rectangular prism.
fgiga [73]

Answer:

  • find a suitable cutting tool
  • cut the prism on the plane of interest

Step-by-step explanation:

A cross section is the intersection of a cut plane with the object of interest. In a classroom setting, we often try to do the cross sectioning mentally or with diagrams, rather than physically cutting anything. Sometimes, there is no substitute for actually performing the cut to see what the cross section looks like.

For certain samples that don't take kindly to cutting, we sometimes encase them in a block of material that helps them hold their shape during the process. Sometimes the "cutting" is performed by grinding away the portion of the material on one side of the cut plane.

7 0
3 years ago
Please help me answer 1-30
Feliz [49]

Answer:

21.) 47/100

Step-by-step explanation:


4 0
3 years ago
The Laplace Transform of a function f(t), which is defined for all t > 0, is denoted by L{f(t)} and is defined by the imprope
lesya692 [45]

(1) D

L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt

Integrate by parts, taking

u = t \implies \mathrm du=\mathrm dt

\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}

Then

L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}

L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}

(2) A

L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}

L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}

7 0
3 years ago
What polygon photo provided
Fed [463]
I may be wrong but I think it would be D, Since (even with the rounded edges) theres only 4 defined sides.
8 0
3 years ago
Read 2 more answers
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