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kifflom [539]
3 years ago
12

Lydia's school is holding a candy fundraiser. They are selling lollipops for $0.50 and candy bars for $1.50 Lydia buys 6 lollipo

ps and some candy bars. If she spends $15.00 in total, how many candy bars, b, did Lydia buy?
Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
8 0

Answer:

8

Step-by-step explanation:

Let's say that the amount of lollipops Lydia buys is represented by x and the number of candy bars is represented by y. For each x, Lydia spends 0.50, and for each y, Lydia spends 1.50. This means that the total amount she spends can be represented by the equation 0.50*x+1.50*y. Our equation is then 0.50*x+1.50*y=15

Using our equation, we can plug 6 in for x, resulting in 0.50*6+1.50*y=15, so 0.50x+1.50y=15\\0.5(6)+1.50y=15\\3+1.50y=15\\1.50y=12\\y=8

In this set of equations, we plugged 6 in, multiplied it out, then subtracted 3 from both sides, and finally divided both sides by 1.50 to get 8 as our answer.

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Step-by-step explanation:

6 0
2 years ago
Z/3 &lt;= 3<br> solve for z
Molodets [167]

Answer:

assuming <= is less than or equal to

then z/3 is less than or equal to 3

z had to be less than 9

4 0
2 years ago
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
4. Bobbie scores m marks in a test. <br> Georgia scores three times as many marks as Bobbie.
xxTIMURxx [149]

Marks of Bobbie = x

Then, marks of Georgia

= 3 times marks of Bobbie

= 3 times m

= 3 × m

= 3m

6 0
2 years ago
Which of the following rates is proportional to $2.50 for 5 cans?
MAVERICK [17]
A, divide 2.5 by 5 and you get 0.5, multiply that by 9 and you get 4.5. A is the answer
4 0
2 years ago
Read 2 more answers
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