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Savatey [412]
3 years ago
7

Jennie has 177 more songs downloaded on her mp3 player than Diamond. Together, they have 895 songs downloaded. How many do each

of them have?
Mathematics
1 answer:
kogti [31]3 years ago
3 0

Answer:jennie has 536 songs and diamond has 359

Step-by-step explanation:

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Solve the following equations using any method: 4x-y=3 -4x+y=-3
harkovskaia [24]
The second equation balances out to -3=-3
And the first one balances out as 3=3 so I have no clue

4 0
3 years ago
Hellppppppppppppppppppppppppp
Aneli [31]

Answer:

126 degrees

Step-by-step explanation:

Every triangle equals to 180 degrees so all you have to do is add 26+28 which equals 54 then subtract that from 180 and there's your answer.

26+28 = 54

180-54 = 126

Hope this helps! :)

7 0
3 years ago
A theme park has a Ferris wheel that takes 2 minutes to go around its 450-foot circumference. The equation · (130) = 450 relates
olchik [2.2K]

Answer: The Ferris wheel’s speed in feet per hour.

Step-by-step explanation:

3 0
3 years ago
Suppose a standard twelve-hour clock now shows a time of 10:45 what time will the clock show in a 100 hour from now
o-na [289]
It should show 2:45. You can use 12 hr increments to get you back to 10:45 and 12 goes into 100 nine times bringing it to 96. Because of that you have 4 more hours to add to 10:45 and that brings you to 2:45.
4 0
3 years ago
Read 2 more answers
Power Series Differential equation
KatRina [158]
The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for y

\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0

which indeed gives the recurrence you found,

a_{n+3}=-\dfrac{n-3}{n+3}a_n

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that a_2=0, and substituting this into the recurrence, you find that a_2=a_5=a_8=\cdots=a_{3k-1}=0 for all k\ge1.

Next, the linear term tells you that 6a_0+6a_3=0, or a_3=a_0.

Now, if a_0 is the first term in the sequence, then by the recurrence you have

a_3=a_0
a_6=-\dfrac{3-3}{3+3}a_3=0
a_9=-\dfrac{6-3}{6+3}a_6=0

and so on, such that a_{3k}=0 for all k\ge2.

Finally, the quadratic term gives 6a_1-12a_4=0, or a_4=\dfrac12a_1. Then by the recurrence,

a_4=\dfrac12a_1
a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1
a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1
a_{13}=-\dfrac{10-3}{10+3}a_{10}=\dfrac{(-1)^3}2\dfrac{7\times4}{13\times10\times7}a_1

and so on, such that

a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}

for all k\ge2.

Now, the solution was proposed to be

y=\displaystyle\sum_{n\ge0}a_nx^n

so the general solution would be

y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots
y=a_0(1+x^3)+a_1\left(x+\dfrac12x^4-\dfrac1{14}x^7+\cdots\right)
y=a_0(1+x^3)+a_1\displaystyle\left(x+\sum_{n=2}^\infty\left(\prod_{i=1}^{n-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\right)x^{3n-2}\right)
4 0
3 years ago
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