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2 years ago

In the Treasure, Treasure! video game, players gain and lose points by finding and losing treasure. In Level 2, Leslie earned

Mathematics

2 answers:

Vanyuwa [196]2 years ago

6 0

So you like stranger things?

USPshnik [31]2 years ago

3 0

Answer:

. finding and losing treasure. In level 2, Leslie earned -22 points when a storm blew a treasure chest off her ship. Then, she earned 34 ... a storm blew a treasure chest off her ship. Then, she earned 34 points for finding more treasure on the deserted island. What is the overall change in Leslie's score during level 2? ansver.

Step-by-step explanation:

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How many words can be formed from the letters of the word 'ENGLISH' ? How many these do not begin with E? How many of these begi

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Total Number of words made out of English = 93

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2 years ago

Who used less sugar 5/10 or 5/6

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It would be 5/6 because the denominator Is smaller than 10

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3 years ago

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Add the two expressions 6q+1 and q+11<br> Enter your answer into the box

In-s [12.5K]

Answer:

7q + 12

Step-by-step explanation:

6q + 1 + q + 11

add like terms. add 6q and q together and then add 1 and 11 together

6q + q + 1 + 11

7q + 12

7q + 12 is your answer. i hope this helps :)

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3 years ago

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

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

The area of a circular garden is 4,225 T square feet. Rick wants to find the diameter of the garden. What is the diameter of

ankoles [38]

Answer:

D. 130 feet

Step-by-step explanation:

The area of a circular garden is 4,225 π square feet

The area of a circle is computed as follows:

Area of a circle = π*D²/4

where D refers to the diameter

Replacing with data:

4,225 π = π*D²/4

4,225 π/π *4 = D²

√16900 = D

130 ft = D

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2 years ago