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2 years ago

What is 35% of 280?

Mathematics

2 answers:

nordsb [41]2 years ago

6 0

Answer:

Step-by-step explanation:

Multiply.38*280

Marrrta [24]2 years ago

5 0

Answer:

Step-by-step explanation:

To find the answer I multiplied 280 by 0.35, that is how you can quickly find a percentage. For example if you would like to find what is 7% of 280 you can just multiply it by 0.07.

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Is negative square root of 35 rational or irrational?

Vika [28.1K]

Irrational would be the correct answer

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2 years ago

The data shown in the table below can be

kvv77 [185]

Answer:

6: 12

7: 14

8: 16

Step-by-step explanation:

You do the last (y) term given [18] minus the first (y) term [10] given divided by how many terms it takes to get from the last (y) term given minus the first (y) term given [4]. So the equation looks like this:

(18 - 10)/4= 8/4 = 2

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2 years ago

Jasmine earns $36 for 4 hours of baby-sitting. She charges a constant hourly rate. Compare the tables. Which correctly shows the

Ymorist [56]

The answer would be A. She charges $9 an hour.

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2 years ago

Read 2 more answers

Calculate the rate of change of the linear function that contains the points

Mandarinka [93]

Answer:

-2

Step-by-step explanation:

X1 Y1 X2 Y2

(-15, 30) (7, -14)

Y2-Y1 = -14-30 = -44 = -2

———————————————

X2-X1 = 7-(-15)= 22 = 1

3 0

2 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,

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

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2 years ago