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3 years ago

4.0 divided by 0.02=

Mathematics

2 answers:

Paha777 [63]3 years ago

8 0

Answer:

4.0/0.02=200

The answer is 200

elena-14-01-66 [18.8K]3 years ago

4 0

Answer:

200

Step-by-step explanation:

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Which fraction is equal to 12%

allochka39001 [22]

Answer:

3/25

Step-by-step explanation:

Simplify the fraction by using common factors of numerator and denominator. Now, 12% can be written as 12/100.

4 0

2 years ago

A rectangle Has an area of 252 yd.². The length of the rectangle is 10 yards longer than twice it’s worth. Find the width and le

Rama09 [41]

Answer:

Width = 9 yds

Length = 28 yds

Step-by-step explanation:

Width = x

Length = 2x + 10

Area is 252 yd²

x(2x + 10) = 252

2x² + 10x - 252 = 0

2 (x + 14) (x - 9) = 0

(x + 14) (x - 9) = 0

x = - 14 or x = 9

Width = 9 yds

Length = 2x9 + 10 = 28 yds

7 0

2 years ago

Read 2 more answers

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

Which equations matches the graph

iren2701 [21]

The answer for the question is C

7 0

2 years ago

Given triangle MNK below, find sin M/ Cos K

meriva

Answer:

The Answer is A

Step-by-step explanation:

its A I got it right

8 0

2 years ago