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Alisiya [41]
3 years ago
15

Which is the graph of an even monomial function?

Mathematics
1 answer:
Cloud [144]3 years ago
4 0

Answer:

The graph of an even function is symmetric about the y-axis. The graph of an odd function is symmetric about the x-axis. It is possible that the use of these two words originated with the observation that the graph of a polynomial function in which all variables are to an even power is symmetric about the y -axis.

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The inequality 10.45b + 56.50 < 292.67 is used to find the number of boxes (b) that can be loaded on a truck without exceedin
PSYCHO15rus [73]
We solve the inequality by subtracting 56.50 from both sides of the equation,
                                10.45b + 56.50 - 56.50 < 292.67 - 56.50 
                                        10.45b < 236.17
Then, divide both sides of the inequality by 10.45
                                             b < 22.6
The solution suggests that the number of boxes than can be loaded on a truck without exceeding the weight limit of the truck should always be lesser than 22.6. Since we are talking about number of boxes, the maximum number of boxes that can be loaded should only be 22. 
6 0
3 years ago
Read 2 more answers
Find the angle whose measure is 5 times its complement
Ivenika [448]

Answer:

answer is 75 degree

Step-by-step explanation:

you will better understand by this

4 0
3 years ago
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Marcus observed that 8 of the 20 games that he had bowled last week were over 180. What percent of his games were over 180?
shepuryov [24]

Answer:

40%

Step-by-step explanation:

Ok so my brain is sorta not working but it said "hey! 20 divided by 2 is 10!"

So I was like ight and divided 80 by 2 to to get 40%.

Hope this helps.

6 0
3 years ago
Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is
strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is 121y''+110y'-24y=0. We propose that the solution of this equations is of the form y = Ae^{rt}. Then, by replacing the derivatives we get the following

121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

121r^2+110r-24=0

Recall that the roots of a polynomial of the form ax^2+bx+c are given by the formula

x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

r_1 = -\frac{12}{11}

r_2 = \frac{2}{11}

So, in this case, the general solution is y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

c_1 + c_2 = 1

c_1\frac{-12}{11} + c_2\frac{2}{11} = 0(or equivalently c_2 = 6c_1

By replacing the second equation in the first one, we get 7c_1 = 1 which implies that c_1 = \frac{1}{7}, c_2 = \frac{6}{7}.

So y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}

b) By using y(0) =0 and y'(0)=1 we get the equations

c_1+c_2 =0

c_1\frac{-12}{11} + c_2\frac{2}{11} = 1(or equivalently -12c_1+2c_2 = 11

By solving this system, the solution is c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}

Then y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}

c)

The Wronskian of the solutions is calculated as the determinant of the following matrix

\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2

By plugging the values of y_1 and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

e^{\int -p(x) dx}

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}

Note that this function is always positive, and thus, never zero. So y_1, y_2 is a fundamental set of solutions.

8 0
3 years ago
The 100th term of ​ 70, 100, 130
fgiga [73]

Answer:

3040

Step-by-step explanation:

given arithmetic progression is

70,100,130,...

here

first term (a)=70

common difference (d)=100-70=30

number of term n=100

using the formula of arithmetic progression

an=a+(n-1)d

a100=70+(100-1)30

a100=70+99×30

a100=70+2970

a100=3040

6 0
2 years ago
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