What is the coefficient of xy^5 in the expansion of (x+y)^6

Mathematics

1 answer:

Savatey [412]3 years ago

7 0

(x+y) is the coefficient of xy

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Write a formula for the general term or nth term for the sequence. Then find the indicated term. five halves comma five fourths

nadya68 [22]

Sequence: 5/2, 5/4, 5/8, 5/16
a8=?

a1=5/2
a2=5/4
a3=5/8
a4=5/16

a2/a1=(5/4)/(5/2)=(5/4)*(2/5)=(5*2)/(4*5)=2/4=1/2
a3/a2=(5/8)/(5/4)=(5/8)*(4/5)=(5*4)/(8*5)=4/8=1/2
a4/a3=(5/16)/(5/8)=(5/16)*(8/5)=(5*8)/(16*5)=8/16=1/2
Ratio: r=a2/a1=a3/a2=a4/a3→r=1/2

an=a1*r^(n-1)
a1=5/2, r=1/2
an=(5/2)*(1/2)^(n-1)
an=(5/2)*[1^(n-1)/2^(n-1)]
an=(5/2)*[1/2^(n-1)]
an=(5*1)/[2*2^(n-1)]
an=5/2^(1+n-1)
an=5/2^n

n=8→a8=5/2^8
a8=5/256

Answers:
The formula for the general term or nth term for the sequence is an=5/2^n
a8=5/256

6 0

3 years ago

I dont understand this at all. I really need help!

Westkost [7]

Hello!

$\large\boxed{x^{4}}$

Recall that:

$\sqrt[z]{x^{y} }$ is equal to $x^{\frac{y}{z} }$ . Therefore:

$\sqrt[3]{x^{2} } = x^{\frac{2}{3} }$

There is also an exponent of '6' outside. According to exponential properties, when an exponent is within an exponent, you multiply them together. Therefore:

$(x^{\frac{2}{3} })^{6} = x^{\frac{2}{3}* 6 } = x^{\frac{12}{3} } = x^{4}$

3 0

3 years ago

Tyler baked 702 cookies he sold them in boxes of 18 after selling all of the boxes of cookies for the same amount of each he ear

jenyasd209 [6]

Keywords:

Cookies, boxes, sale, profit, unit cost

For this case we have to bake an amount of 702 cookies, which were distributed in boxes of 18 for sale. Selling each box at the same price earned a profit of 136.50 dollars, you want to know the unit cost of each box of cookies.

Let $"x"$ be the number of boxes of cookies:

So, to know the number of boxes we divide 702 cookies among 18:

$x = \frac {702} {18}\\x = 39$

So, we have Tyler sold 39 boxes of cookies

Now, let $"y"$ be the cost of each box of cookies, to know the unit cost or sale price of the cookie boxes, we divided 136.50 dollars among 39 boxes of cookies, then: