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Anestetic [448]
2 years ago
15

Last question please help!

Mathematics
1 answer:
viktelen [127]2 years ago
4 0

Answer:

Angle BFC.

Step-by-step explanation:

Because angle BFD is a right angle because angle AFE is equal to BFD because of vertical angles

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Describe how the Commutative and Associative Properties of Addition can make adding mixed numbers easier.
Sloan [31]
Because both properties just switch around, and you get the same answer
3 0
3 years ago
Read 2 more answers
Y = -(x+5)2 + 9 into standard form
mote1985 [20]

Answer:

y = -2x - 1

Step-by-step explanation:

y = -(x+5)2 + 9

Foil method to get rid of parenthesis

y= -2x - 10 + 9

y = -2x - 1

Have a great day!

(Please mark this question as brainiest if it worked or ask me a question if you do not understand)

4 0
2 years ago
Select all that apply.<br> Which words describe the square root of 100?
abruzzese [7]

The square root of a number is any number that when multiplied against itself becomes the number under symbol denoting for the square root operation.

The number 10 is the result of the square root operator applied to the number 100 because 10 * 10 = 100.

The number 10 is a natural number because natural numbers are positive integers. It is also an integer, whole number, and rational.

Hope this answer helps!

5 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
Prove or disprove that the product of two irrational numbers is irrational
DanielleElmas [232]
Hi. The statement you gave is somewhat true, but not always. I disprove this and there are tons of websites that can help you with more in depth explanation. Have a good day.
6 0
3 years ago
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