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3 years ago

How to work ratios and graphs

Mathematics

2 answers:

Olegator [25]3 years ago

7 0

Do you have a picture?

Mila [183]3 years ago

7 0

I'm not sure about ratios, because I'm bad at them myself, but I can help you with graphs. Take this plot point, for example (3,4) The X-axis is the line that is horizontal. This is the first letter in the coordinates, 3, so you would find the number 3 on the X-axis. The Y-axis is the line that is vertical, and its number is the second one in your set of coordinates. You would then find 4 on your Y-axis. Then, draw a line from your X-coordinate up, and your Y-coordinate right. Find where these overlap. If these are negatives, however (for example, if the coordinates were (-3,-4)), then draw the line in the opposite direction (X-coordinate down, Y-coordinate left) and do the same as you would with normal coordinates and find where they overlapped. I hope this helped you, since I love graphs and want to help!

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How to make my answer for 0.70 a fraction

miss Akunina [59]

Answer:

0.70 = 7/10

Step-by-step explanation:

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2 years ago

Read 2 more answers

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

2 years ago

19. Which is the suitable identity used to evaluate 9992 ?

andrew11 [14]

Answer:

I suppose you mean 999^2

(1000-1)^2

which comes under the identity

(a-b)^2=

(a^2 + b^2 - 2ab)

hope this helps

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2 years ago

What tells us how many steady beats go in a measure?

Step2247 [10]

Answer:

You'll spot the time signature in the beginning of the music – it's two numbers stacked vertically. The top number tells you how many beats there are in one measure.

Step-by-step explanation:

hope it helps

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1 year ago

Dion went out to lunch. The bill, including tax, was $30. Dion left a tip of $6. What percent of the bill did Dion leave as tip?

KonstantinChe [14]

The percentage would be 20%

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3 years ago