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4 years ago

11

Write an inequality for the situation to pay a child's price you must be 12 years or younger why is there not only one correct a

nswer to an inequality

1 answer:

Alex_Xolod [135]4 years ago

3 0

5x+2<4×+4 this inequality will pay a childs price which is 2 $

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Which equation below represents a line with a slope of - 2/3 that passes through the point (-4.5 , 10)?

Lostsunrise [7]

Answer:

Option C

Step-by-step explanation:

If a line passes through a point, then that point is called a solution to the line's equation. Substituting the x and y values of that solution into the equation will give a true statement. So, to find out which option is correct, we can substitute the x and y values of (-4.5, 10) into each equation and see if the result is a true statement.

Let's try this with option C. To make things easier, convert -4.5 into decimal form: $-4\frac{1}{2}$ . Substitute $-4\frac{1}{2}$ for x and 10 for y in the equation, then solve:

$y = -\frac{2}{3} x+7\\\\10 = -\frac{2}{3} (-4\frac{1}{2} )+7\\10 = -\frac{2}{3} (-\frac{9}{2})+7\\10 = 3+7\\10 = 10$

10 does equal 10, so this is a true statement. Option C is the right answer.

8 0

3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago

4x/3 + 2(3- x) = 5 Solve for x as an FRACTION

forsale [732]

Answer:

<h2>The answer is .... <u><em> x = 3/2</em></u><em> </em>and if you want to write it as a mixed fraction i will be x= <u><em>1 and 1/2</em></u></h2>

Step-by-step explanation:

Hope this Helped :)

6 0

3 years ago

Help me plz I need the answer asap

PilotLPTM [1.2K]

Answer:

The grandmother is who to go to for heavy metal. The grandmother has a 60% chance of getting something else while the grandfather has a 71% chance. The odds aren't good but granny is who to go with.

6 0

3 years ago

Select the four relations that represent a function.

Anton [14]

The first, third, fifth and sixth one are all functions.

8 0

3 years ago