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3 years ago

Help meeeee!!!!!!! I’ll give you a cookie.

Mathematics

1 answer:

lianna [129]3 years ago

7 0

Answer:

(-5,24)

Step-by-step explanation:

2f+3g-h

first we substitute in the numbers

2(-3)+3(1)-2

-6+3-2

-3-2

-5

Then again

2(5)+3(4)-(-2)

10+12+2

22+2

hope this helps!:)

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2 years ago

Fine the product mentally. [(x+1) (x-1)^2

Murljashka [212]

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5 0

3 years ago

Read 2 more answers

A sequence is defined by the recursive function f(n + 1) =1/3 f(n). if f(3) 9= , what is f(1)

gayaneshka [121]

Answer:

f(1) = 81

Step-by-step explanation:

f(n + 1) = 1/3 f(n)

⇔ f(n) = 3 × f(n + 1)

……………………………

if f(3) = 9 ⇒ f(2) = 3 × f(3) = 3 × 9 = 27

Then

f(1) = 3 × f(2) = 3 × 27 = 81

5 0

2 years ago

Find the tenth term of the expansion (x+ y)¹³

Anon25 [30]

Answer:

$715x^4y^9$

Step-by-step explanation:

Given

$(x + y)^{13$

Required

Determine the 10th term

Using binomial expansion, we have:

$(a + b)^n = ^nC_0a^nb^0 + ^nC_1a^{n-1}b^1 + ^nC_2a^{n-2}b^2 +.....+^nC_na^{0}b^n$

For, the 10th term. n = 9

So, we have:

$(a + b)^n = ^nC_{9}a^{n-9}b^{9$

$(x + y)^{13} = ^{13}C_{9}x^{13-9}y^9$

$(x + y)^{13} = ^{13}C_{9}x^4y^9$

Apply combination formula

$(x + y)^{13} = \frac{13!}{(13-9)!9!}x^4y^9$

$(x + y)^{13} = \frac{13!}{4!9!}x^4y^9$

$(x + y)^{13} = \frac{13*12*11*10*9!}{4!9!}x^4y^9$

$(x + y)^{13} = \frac{13*12*11*10}{4!}x^4y^9$

$(x + y)^{13} = \frac{13*12*11*10}{4*3*2*1}x^4y^9$

$(x + y)^{13} = \frac{17160}{24}x^4y^9$

$(x + y)^{13} = 715x^4y^9$

Hence, the 10th term is $715x^4y^9$

3 0

3 years ago