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1 year ago

The function h is defined by the following rules

Mathematics

1 answer:

wolverine [178]1 year ago

3 0

Answer:

1, 2, 4, 5, 7

Step-by-step explanation:

The function rule tells you what you must do to each input (x) value in order to find the value of the function (h(x)).

The function rule here tells you that 2 is added to the x-value.

For example, for x = -1, ...

h(x) = x +2 . . . . . . given function rule

h(1) = -1 +2 = 1 . . . . . . put the value of x where x is, and do the arithmetic

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Which are solutions of x2 + 5x = 2? Check all that apply.

ANEK [815]

we have

$x^{2} +5x=2$

Complete the square. Remember to balance the equation by adding the same constants to each side

$x^{2} +5x+2.5^{2}=2+2.5^{2}$

$x^{2} +5x+6.25=2+6.25$

$x^{2} +5x+6.25=8.25$

Rewrite as perfect squares

$(x+2.5)^{2}=8.25$

Square Root both sides

$(x+2.5)=(+/-)\sqrt{8.25}$

$x=(+/-)\sqrt{8.25}-2.5$

$\sqrt{8.25}=\frac{\sqrt{33}}{2}$

substitute

$x=(+/-)\frac{\sqrt{33}}{2}-2.5$

therefore

<u>the answer is</u>

the solutions are

$x1=-2.5+\frac{\sqrt{33}}{2}$ or $x1=\frac{-5+\sqrt{33}}{2}$

$x2=-2.5-\frac{\sqrt{33}}{2}$ or $x2=\frac{-5-\sqrt{33}}{2}$

7 0

3 years ago

Read 2 more answers

At the start of the month the value of the investment is 48.45. By the end of the month the value investment changed by a loss o

Ket [755]

Value of the investment (loss of 13.8dollars)
48.48-13.8=34.65dollars

7 0

4 years ago

Find the domain and the range of the relation shown on the

irinina [24]

The domain of a graph is the possible values of x, the graph can take.

<em>(b) The domain of the relation is the interval [-10,10]</em>

From the attached graph, we have the following observations on the x-axis.

<em>The value of x starts from -10</em>
<em>The value of x ends at 10</em>

So, the domain of x is from -10 to 10

Using interval notation, the domain of the relation is: $[-10,10]$

Read more about domains at:

brainly.com/question/16875632

5 0

2 years ago

Simplify the given expression, using only positive

zalisa [80]

Answer: x=4 y=14 z=4

Step-by-step explanation:

On edge it is that

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

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