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chubhunter [2.5K]
4 years ago
11

A 3pack of boxes of juice costs 1.09. A 12 pack of boxes cost 4.49. A case of 24 boxes costs 8.78. Which is the best buy?Explain

your reasoning
Mathematics
1 answer:
Leona [35]4 years ago
5 0
1.09÷3= .3633 each
4.49÷12= .37416 each
8.78÷24= .3658 each

so the first one
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I need help on this I forgot how to do the work, answer please?
san4es73 [151]
The measure of the right angle is 90 degrees. add 25 to 90 to get 115. 180-115= 65 so your answer is 65
8 0
3 years ago
How does the shape of a cross-section of a 3D figure parallel to the base relate to the volume of the 3D figure?
Ket [755]

Answer:

The shape of each cross-section of a 3D figure,  relates to the volume because the area of the cross-section is determined by its shape and the area of this cross section is in the sum that calculates the volume of this 3D figure.

Step-by-step explanation:

An infinite sum of all the all the cross-sections of a 3D figure parallel to the base equals the volume of that 3D figure.

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3 years ago
How does the area of the triangle relate to the area of the square? Please Explain!!! I really don't get it!
kolbaska11 [484]
A square has 4 right angles and 4 equal sides.
If you draw a diagonal through the square then you get two right triangles. So the area relation between a square and a right triangle is going to be half. The area of the triangle is half the area of the square. 

Also the formula for area of a square is length times width.
And the formula for the area of a right triangle is 0.5 times length times width.

So if you set up the equation (let's use * as multiplication symbol ): 

Area of a Square : Area of a Right Triangle
length * width : 0.5 * length * width
(Divide by length * width on both sides)
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4 0
3 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
Which equation can be used to solve for x in the following diagram.
Romashka [77]

Answer:

11, A

Step-by-step explanation:

90 - 35 = 55

55/5 = 11

6 0
3 years ago
Read 2 more answers
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