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Blizzard [7]
3 years ago
6

PLZ SOME ONE HELPS QUICK ILL GIVE 50 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
neonofarm [45]3 years ago
8 0
We already know that the distance of all the planets are generally calculated by keeping the Sun as the main location point.
Explanation:
The distances of all the planets from the Sun in scientific notation and exponential form-
Mercury-
57
million kilometers.
Scientific notation-
5.7
⋅
10
7

km


Venus-
108
million kilometers.
Scientific notation-
1.08
⋅
10
8

km


Earth-
150
million kilometers
Scientific notation-
1.5
⋅
10
8

km


Mars-
228
million kilometers
Scientific notation-
2.28
⋅
10
8

km


Jupiter-
779
million kilometers
Scientific notation-
7.79
⋅
10
8

km


Saturn-
1.43
billion kilometers
Scientific notation-
1.43
⋅
10
9

km


Uranus-
2.88
billion kilometers
Scientific notation-
2.388
⋅
10
9

km


Neptune-
4.5
billion kilometers
Scientific notation-
4.5
⋅
10
9

km
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sergejj [24]
The answer is 43 because 36 + 7 = 43
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<span>2(a+b)=c
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3 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
2 years ago
Can someone help me with this one math problem ? picture is below
Scorpion4ik [409]
I am not sure on how to do this however here is a link with information thst will be able to help you around this subject.i hope this helps
https://www.bbc.com/bitesize/guides/z8k887h/revision/1
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e-lub [12.9K]

Answer:

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Step-by-step explanation:

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