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yawa3891 [41]
2 years ago
14

Two cylinders, a and b, each started with different amounts of water. The graph shows how the height of the water changed as the

volume of water increased in each cylinder. Which cylinder has a larger radius? Explain how you know.
Mathematics
2 answers:
Sauron [17]2 years ago
7 0

Answer:

:)

Step-by-step explanation:

Elis [28]2 years ago
5 0

Answer:

There is no graph

Step-by-step explanation:

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Solve the quadratic equation for x. What is the product when the roots are multiplied? (x + 2)2 = 1 A) −3 B) −6 C) −9 D) 3
ehidna [41]

Answer:

D

Step-by-step explanation:

To solve the quadratic equation, use the inverse operation to squaring. This will undo the exponent and allows x to be isolated. The inverse operation is the square root. This is called the square root property of equality.

(x + 2)² = 1

√(x+2)² = √1

(x+2) = ±1

x = 1 -2     and   x = -1 - 2

x = -1        and   x = -3

The product of the roots -1 and -3 is -1*-3 = 3

8 0
2 years ago
A baseball player had 4 hits in 8 games. At this rate, how many hits will the baseball player have in the next 28
klemol [59]

Answer:14

Step-by-step explanation:

8 divided by 2 = 4 hits

So

28 divided by 2 = 14

5 0
3 years ago
A train was traveling at a constant speed. The table below shows the distance, in miles, the train traveled the first 4 hours, W
myrzilka [38]

Answer:

need please

Step-by-step explanation:

it's for today please help me

5 0
2 years ago
Identifying the equivalent expression
iragen [17]

Answer:

\large\boxed{\sqrt{x+3}=(x+3)^\frac{1}{2}}

Step-by-step explanation:

\text{Use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\sqrt{x+3}=\sqrt[2]{x+3}=(x+3)^\frac{1}{2}

6 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
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