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3 years ago

20 dollars was one quarter of the money spent on the pizza

Mathematics

1 answer:

NNADVOKAT [17]3 years ago

3 0

Answer:

$20.00 was one-quarter of the money spent on pizza.

14m=20.00 What divided by 4 equals 20.00?

The solution is 80. So, the amount of money spent on pizza was $80.00.

By working through this question mentally, you were applying mathematical rules and solving for the variable m.

Step-by-step explanation:

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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 $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

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

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 $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

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

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The area of the triangle is 22.4cm^2.

Ivanshal [37]

Answer:

The longest side length b is approximately 13.172 cm

Step-by-step explanation:

The given parameters are;

The area of the triangle = 22.4 cm²

The given angles of the triangle = ∠97°

The given side of the triangle = 3.7 cm

The formula for the area of a triangle, A = 1/2 × a × b × sin(C)

Where;

a, and b are the side legs forming the angle C

Therefore, we have;

A = 22.4 = 1/2 × 3.7 × a × sin(97°)

a = 22.4/(1/2 × 3.7 × sin(97°)) ≈ 12.199

a ≈ 12.199 cm

Therefore, by cosine rule, we have;

b² = a² + c² - 2 × a × c × Cos(B)

Substituting the values, gives;

b² = 12.199² + 3.7² - 2 × 12.199 × 3.7 × Cos(97°)

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