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Alex777 [14]
3 years ago
6

Which table represents a linear function?

Mathematics
2 answers:
Anettt [7]3 years ago
8 0
Where are the pictures Nd tables?
Vesna [10]3 years ago
3 0
There are no pictures or tables
You might be interested in
How to find the range when the domain is already given.
Whitepunk [10]
You find the range by adding all the sides I think
5 0
3 years ago
Find a solution to the initial value problem, y′′+18x=0,y(0)=5,y′(0)=1.
Serga [27]

We want to find a solution to the initial value problem:

y'' + 18x = 0 \qquad,\qquad y(0) = 5 \qquad,\qquad y'(0)=1.

We can start by integrating the equation once:

\dfrac{\textrm{d}^2 y}{\textrm{d}x^2} + 18 x = 0 \iff \dfrac{\textrm{d}^2 y}{\textrm{d}x^2} = -18 x \iff\\\\\iff \dfrac{\textrm{d}y}{\textrm{d}x} = -18\displaystyle\int x\textrm{ d}x \iff \dfrac{\textrm{d}y}{\textrm{d}x}=-18\dfrac{x^2}{2} + C \iff\\\\\iff \dfrac{\textrm{d}y}{\textrm{d}x} = -9x^2 + C.

Using the initial condition y'(0) = 1, we can determine the integration constant C:

\dfrac{\textrm{d}y}{\textrm{d}x}\Big\vert_{x= 0} = 1 \iff -9 \times 0^2 + C = 1 \iff C = 1.

Therefore, we have:

\dfrac{\textrm{d}y}{\textrm{d}x} = -9x^2 + 1

We can now integrate again:

y(x) = \displaystyle\int\dfrac{\textrm{d}y}{\textrm{d}x}\textrm{ d}x = \int\left(-9x^2+1\right)\textrm{d}x = -9\int x^2\textrm{ d}x + \int\textrm{d}x =\\\\= -9\dfrac{x^3}{3} + x + K = -3x^3 + x + K.

The integration constant K is determined by using y(0) = 5:

y(0) = 5 \iff -3 \times 0^3 + 0 + K = 5 \iff K = 5.

Finally, the solution is:

\boxed{y(x) = -3x^3 + x + 5}.

7 0
3 years ago
What is the amplitude of sin ?
sp2606 [1]

You haven't provided a graph or equation so I will tell the simplified meaning of amplitude instead.

Amplitude, is basically a distance from midline/baseline to the maximum or minimum point.

For sine function, can be written as:

\displaystyle \large{ y = A \sin(bx  -  c) + d}

  • A = amplitude
  • b = period = 2π/b
  • c = horizontal shift
  • d = vertical shift

I am not able to provide an attachment for an easy view but I will try my best!

We know that amplitude or A is a distance from baseline/midline to the max-min point.

Let's see the example of equation:

\displaystyle \large{y = 2 \sin x}

Refer to the equation above:

  • Amplitude = 2
  • b = 1 and therefore, period = 2π/1 = 2π
  • c = 0
  • d = 0

Thus, the baseline or midline is y = 0 or x-axis.

You can also plot the graph on desmos, y = 2sinx and you will see that the sine graph has max points at 2 and min points at = -2. They are amplitude.

So to conclude or say this:

If Amplitude = A from y = Asin(x), then the range of function will always be -A ≤ y ≤ A and have max points at A; min points at -A.

6 0
3 years ago
There are thirteen animals in a barn. Some are chickens and some are pigs there are 40 legs in all. How many of each animal are
slega [8]
Hi. In order to solve this word problem, you must make two equations and solve them together. To make it easy, we will let the letter "p" represent the pigs and the letter "c" for the chickens. So our first equation will be:

p + c = 13

Now to count the legs - we know that pigs have 4 legs and chickens have 2 so our next equation will be:

2c + 4p = 40

Using the first equation we see that c = 13 - p Now we insert this into the second equation and solve:

2x(13-p) +4 pm = 40
26 - 2p + 4p = 40
2p = 14
p = 7 (there are 7 pigs so now we know there are 6 chickens.

Answer: 7 pigs and 6 chickens

I hope this helps.

Take care,
Diana
3 0
3 years ago
(15 Points)
expeople1 [14]

ANSWER 1



Note that,


f(u)=tan^{-1}(u)


is the same as


f(u)=arctan(u)



We apply the product rule.


f(x)=x^2tan^{-1}(x)


So we keep the second function and differentiate the first,plus we keep the first function and differentiate the second.


f'(x)=(x^2)'tan^{-1}(x)+x^2(tan^{-1}(x))'



Recall that,

If

f(u)=tan^{-1}(u)



Then,

f'(u)=\frac{1}{1+u^2}} \times u'


This implies that,

f'(x)=2xtan^{-1}(x)+\frac{x^2}{x^2+1}



ANSWER 2


We apply the product rule and the chain rules of differentiation here.



f(x)=xsin^{-1}(1-x^2)




f'(x)=x'sin^{-1}(1-x^2)+x(sin^{-1}(1-x^2))'



Recall that,

If

f(u)=sin^{-1}(u)



Then,

f'(u)=\frac{1}{\sqrt{1-u^2}} \times u'



This implies that,


f'(x)=sin^{-1}(1-x^2)+x \times \frac{1}{\sqrt{1-(1-x^2)^2}}\times (-2x)


f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-(1-2x^2+x^4)}}


f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-1+2x^2-x^4}}



f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{2x^2-x^4}}





5 0
4 years ago
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