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3 years ago

17,37 + 0,53x7= tolong pls

Mathematics

2 answers:

Nutka1998 [239]3 years ago

7 0

2108
The answer is D

Explain: used a Calculator

grin007 [14]3 years ago

6 0

Answer:

21,08

Step-by-step explanation:

3.71+17.37=21.08

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(2.3 x 108)/(9.8 x 108)

QveST [7]

Answer: 9.45

Step-by-step explanation:

2.3 * 108 = 248.4

9.8 * 108 = 1,058.4

248.4 divided by 1,058.4 = 9.45

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3 years ago

Anybody want to help with math? Pls hurry, I’ll give brainiest and everything.

Vsevolod [243]

Answer:

Step-by-step explanation:

You are going to add 9 to 27 because you are counting by nines on the bottom.

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2 years ago

Which is the simplified form of the expression (6^-2•6^5)^-3

Arisa [49]

Answer:

Step-by-step explanation:

$( {6}^{ - 2} \times {6}^{5} ) ^{ - 3} \\ = ( {6}^{3} )^{ - 3} \\ = {6}^{0} \\ = 1$

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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

3 years ago

What problem types can be solved using law of sines?

aev [14]

Let's take a triangle ABC, with a, b, and c the sides length, he law of sine is:

a/sin A =b/sin B = c/sin C
If we know the value of 2 angles and one side or the value of 2 sides and one angle, we can calculate all the elements of the triangle

6 0

3 years ago

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17,37 + 0,53x7= tolong pls​

17,37 + 0,53x7= tolong pls