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3 years ago

What is 2/5 x 1/4 Then Simplified Expression?

Mathematics

1 answer:

weeeeeb [17]3 years ago

6 0

Answer:

2/20 = 0.1

Step-by-step explanation:

2/5 × 1/4
2×1/5×4
2/20

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37% of what number is 99.9?

siniylev [52]

Answer:

270

Step-by-step explanation:

37 % of a number is the same as divide by 100 and multiply by 37. To get this number we do the opposite: divide 99.9 by 37 which gives us 27 and then multiply by 100 which will give us 270.

if we check this: 270 /100 *37 = 99,9

3 0

3 years ago

MICE is a rhombus. Solve for each missing angle measure.<br><br> <1<br> <2<br> <3

spayn [35]

Answer:

the answer is :

<1=90°

<2=41°

<3=49°

5 0

2 years ago

What is 1.65 as a fraction

marshall27 [118]

33/20 as well as 1 and 13/20. First you look it up in the internet aka Google.

4 0

2 years ago

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

What’s the answer???

stellarik [79]

Answer: (3,4)

Each input can only have one output for a function

The other pairs give a different output for the input so that would make it a relation not a function.

7 0

3 years ago