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2 years ago

Which snail travels the fastest

Mathematics

2 answers:

Nadusha1986 [10]2 years ago

6 0

Answer:

Step-by-step explanation:

ANTONII [103]2 years ago

3 0

There’s no picture in the present

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Write 0.6 as a fraction ?

satela [25.4K]

Answer:

6/10 ; 3/5 (simplified form)

Step-by-step explanation:

To solve, note the decimal place value. Move the decimal point to the right two place values, and set it over 100

0.6 = 60/100

Simplify the fraction. Divide common factors from the numerator & denominator (20 in this case)

(60/100)/(20/20) = 3/5

3/5 is your answer

7 0

3 years ago

Read 2 more answers

PLEASE HELP Residents of Eastport are convinced about the number of drivers who speed when driving down Main street. A group of

ZanzabumX [31]

Answer:

The answer is B I think

Step-by-step explanation:

They are talking about how they are convinced about the people who speed

3 0

2 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

3 years ago

Please help! Will mark brainliest!

Anestetic [448]

Each spinner has 5 numbers, the total combinations would be 5 x 5 = 25 combinations,

The answer is 25

6 0

3 years ago

Whitch expression is equivalent to 7 to the 3rd power? 7*7*7. Or 21. Or. 3 to the power of. Or. 10

laila [671]

Answer:

7x7x7

Step-by-step explanation:

8 0

2 years ago

Which snail travels the fastest​

Which snail travels the fastest