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mr Goodwill [35]
2 years ago
5

105.25 x 70.75 NO LINKS!!!!!

Mathematics
1 answer:
Anarel [89]2 years ago
3 0

Answer:

7446.4375

Step-by-step explanation:

  105.25

x   70.75

-------------

 52625

First, we will want to start with the hundredths place. multiply 5×5. We get 25. Put the 5 down and carry the 2 above 2 in 105.25. Next, we multiply 2 in 105.25 with 5 in 70.75. 2×5 is 10, then we add the carried 2, and 10+2 is 12. We put the 2 down below, and carry the 1 above 5 in 105.25. Next, we multiply the 5 in 105.25 times 5. 5×5 is 25. We then need to add the carried 1 to 25 to get 26. We put the 6 down and put the 2 over 0 in 105.25. Next, we multiply 0 in 105.25 with 5 in 70.75. 0×5 is 0, plus our carried 2 from before, so 2. We don't need to carry anything since it's only the ones digit, so we put the 0 down. Next, we multiply the 1 in 105.25 with 5 again in 70.75. 1×5=5. Since we have no carries we can put the 5 down without adding anything. If this seemed confusing, let me simplify it for you. You basically start with the bottom number's "digit". For us, that's 70.75, and 5 would be the "digit." (The last digit from the bottom number.) We then multiply this "digit" by every number from the top number's digits. For example, we'd multiply 70.75's 5, with 105.25's 5 first, then continue on going left from there. I hope this helped! If it just confused you more please let me know.  (Also remember to put a 0 when you start multiplying with your second digit on the bottom number)

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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
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