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katrin [286]
2 years ago
5

What is the domain of f(x) = 5^x?

Mathematics
2 answers:
Natali5045456 [20]2 years ago
8 0
Your domain is B, all real numbers.  That's an exponential growth function.The 5 just makes it a steeper graph, closer to the y axis, that's all.
Tju [1.3M]2 years ago
7 0

Answer:

D.

All real numbers

Step-by-step explanation:

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You have at most $3.65 to make copies for a school project. Each copy costs $0.25. Write and solve an inequality that represents
slamgirl [31]

The equation for the problem is 0.25x ≤ 3.65

Hope it helps!

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3 years ago
11, 16, 21, ...<br> Find the 44th term.<br> PLZ HELP
VashaNatasha [74]

answer : 226

difference : 5

‘0th’ term : 6

rule : 5n + 6

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4 0
3 years ago
Read 2 more answers
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
2 years ago
How do u solve this problem
Feliz [49]
It is a 3:1 ratio of male car drivers to female truck drivers
4 0
3 years ago
Please Help.<br> Write the equation of linear Inequalities to represent each graph.
DerKrebs [107]

Green Line:

This is a horizontal line at x = 0, so it would be:

x ? 0      

Since the shaded area is going to the left of 0 (this means x is negative):

x ≤ 0


Blue Line:

The shaded area is below the line, so:

y ≤ mx + b

When x = 0, y is -5.

y ≤ mx - 5

From each point, you go up 3 units, and to the right 4 units, so the slope is \frac{3}{4}.


y\leq \frac{3}{4}x-5

7 0
3 years ago
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