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denis23 [38]
3 years ago
6

How do i do 3 part a ?

Mathematics
1 answer:
WINSTONCH [101]3 years ago
6 0


In general the binomial expansion is


(a+b)^n = {n \choose 0} a^0 b^n + {n \choose 1} a^1 b^{n-1} + {n \choose 2} a^2 b^{n-2} + ... + {n \choose n} a^n b^0


So in our case, because we want ascending powers of x we'll write,


(-3x + 1)^{11} =  {11 \choose 0} (-3x)^0 1^{11} + {11 \choose 1} (-3x)^{1} 1^{10} + {11 \choose 2} (-3x)^{2} 1^9  + {11 \choose 3 } (-3x)^3 1^8 + ...


We need to calculate the binomial coefficients:


{11 \choose 0}  = 1


{11 \choose 1}  = 11


{11 \choose 2}  = \dfrac{11 \times 10}{2} = 55


{11 \choose 3}  = \dfrac{11 \times 10 \times 9}{3 \times 2} = 165


(-3x+1)^{11} =  1 (-3x)^0 1^{11}  + 11(-3x)^{1} 1^{10}  + 55 (-3x)^2 1^{9}  + 165 (-3x)^3 1^8 + ...


(1-3x)^{11} =  1 -33 x + 495 3x^2 - 4455 x^3+ ...



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Plz help u have to do all 3 parts
Jobisdone [24]

Answer:

all you have too do is 750*750 = 562,500

Step-by-step explanation:

i hope this helps and pleas give me a stare and a heart thank you so much if you do and if you do not still thank you

7 0
2 years ago
A fruit stand has 28 apples, 20 oranges, and 12 pears. What is the ratio of apples to the total pieces of fruit?
ruslelena [56]
Data:
Apples = 28
Oranges = 20
Pears = 12

<span>What is the ratio of apples to the total pieces of fruit?
Solving:
</span>\frac{apples}{apples+oranges+pears} =  \frac{28}{28+20+12} =  \frac{28}{60} \frac{\div4}{\div4} \to\:\boxed{\boxed{ratio = \frac{7}{15}}}\end{array}}\qquad\quad\checkmark<span>
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2 years ago
Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located
arsen [322]
Check the picture below.

\bf \textit{using the pythagorean theorem}&#10;\\\\&#10;c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\&#10;-------------------------------

\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~  cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~  tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}&#10;\\\\\\&#10;csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}

3 0
3 years ago
Consider the following sequence
wlad13 [49]

Answer:

a₇ =  2.375

Step-by-step explanation:

There is a common ratio r between consecutive terms, that is

r = \frac{-76}{152} = \frac{38}{-76} = \frac{-19}{38} = - \frac{1}{2}

This indicates the sequence is geometric with nth term

a_{n} = a₁ (r)^{n-1}

where a₁ is the first term and r the common ratio

Here a₁ = 152 and r = - \frac{1}{2} , then

a_{n} = 152 (-\frac{1}{2}) ^{n-1} , so

a₇ = 152 (-\frac{1}{2}) ^{6} = 152 × \frac{1}{64} = 2.375

3 0
3 years ago
Read 2 more answers
16. Which of the following are equivalent to(x) 16xA) g(r) 8.21B) g(c) 4096.16-3g (x) -4.4xD) g(x) 0.0625-16'+1E) g(x) 32.16E) g
swat32
F(x) = 16ˣ

A. g(x) = 8(2ˣ)
    g(x) = (2³)(2ˣ)
    g(x) = 2ˣ⁺³
The answer is not A.

B. g(x) = 4096(16ˣ⁻³)
    g(x) = (16³)(16ˣ⁻³)
    g(x) = 16ˣ
The answer is B.

C. g(x) = 4(4ˣ)
     g(x) = 4ˣ⁺¹
The answer is not C.

D. g(x) = 0.0625(16ˣ⁺¹)
     g(x) = (16⁻¹)(16ˣ⁺¹)
     g(x) = 16ˣ
The answer is D.

E. g(x) = 32(16ˣ⁻²)
    g(x) = (2⁵)(2⁴ˣ⁻⁸)
    g(x) = 2(⁴ˣ⁻³)
The answer is not E.

F. g(x) = 2(8ˣ)
    g(x) = 2(2³ˣ)
    g(x) = 2³ˣ⁺¹
The answer is not F.

The answer is B and D.
6 0
2 years ago
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