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2 years ago

Please help with two math questions!!

Mathematics

2 answers:

Brrunno [24]2 years ago

6 0

Their is no questions

IceJOKER [234]2 years ago

3 0

I am not able to understand the question

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What is the value of the expression when n=3 -2n(5+n-8-3n)

igor_vitrenko [27]

-2(3)(5+3-8-3(3))
=-6(5+3-8-9)
=-30-18+48+54
=54

4 0

2 years ago

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77 is what percent of 149

Alexus [3.1K]

77 percent of 149 is 114.73

5 0

3 years ago

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Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

2 years ago

In the term 4p, the variable p is raised to the power of 0 or 1 ?

Svetlanka [38]

Answer:

Step-by-step explanation:

anything raised to the power of 0 would equal 1 and p doesn't equal 1 here.

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2 years ago

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The landscaper pours 200 gallons of herbicide in a pond. The herbicide degrades 10% each week. The landscaper will put another d

wel

8weeks
10% of 200 is 20if you subtract 20 from 200 then you have to do it 8 times to get below 50
#1) 200-20=180
#2) 180-20=160
#3) 160-20=140
#4) 140-20=120
#5) 120-20=100
#6) 100-20=80
#7) 80-20=60
#8) 60-20=40

5 0

3 years ago