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4 years ago

Jupiter has a mass of 1,898,000,000,000,000,000,000,000,000 kg . How would this number be expressed in scientific notation

Mathematics

1 answer:

Elena-2011 [213]4 years ago

3 0

Answer:

1.898 × 10^27

Step-by-step explanation:

We move the decimal point to the left between 1 and 8

and that is 27 places so you multiply the number you get by 10^27

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Solving by substitution

Lesechka [4]

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3 years ago

Read 2 more answers

What is the approximate area of a 70 decree sector of a circle with a radius of 8 inches

puteri [66]

The area of the sector will also be 70/360
Sector area = 70/360 • pir^2
= 7/36 • pi • 8^2
A= 112/9 • pi
A= 39.1 inch^2

3 0

3 years ago

Question -

Sauron [17]

$\rule{200}4$

Answer : The required ratio is (14m-6):(8m+23) .

$\rule{200}4$

Here we are given that the ratio of sum of first n terms of two AP's is (7n + 1):(4n + 27) .

That is.

$\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{7n +1}{4n +27} \\$

As , we know that the sum of n terms of an AP is given by ,

$\small\sf\longrightarrow \pink{ S_n =\dfrac{n}{2}[2a +(n-1)d]} \\$

Assume that ,

First term of 1st AP = a
First term of 2nd AP = a'
Common difference of 1st AP = d
Common difference of 2nd AP = d'

Using this we have ,

$\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{\dfrac{n}{2}[2a + (n-1)d]}{\dfrac{n}{2}[2a' +(n-1)d'] } \\$

$\small\sf\longrightarrow \dfrac{7n+1}{4n+27}=\dfrac{2a + (n -1)d}{2a' + (n -1)d' } . . . . . (i) \\$

Now also we know that the nth term of an AP is given by ,

$\longrightarrow\sf\small \pink{ T_n = a + (n-1)d}\\$

Therefore,

$\longrightarrow\sf\small \dfrac{T_{m_1}}{T_{m_2}}= \dfrac{ a + (n-1)d }{a'+(n-1)d'}. . . . . (ii)\\$

$\longrightarrow\sf\small \dfrac{T_1}{T_2}=\dfrac{2a + (2n-2)d}{2a'+(2n-2)d'} . . . . . (iii)\\$

From equation (i) and (iii) ,

$\longrightarrow\sf\small n-1 = 2m-2\\$

$\longrightarrow\sf\small n = 2m -2+1 \\$

$\longrightarrow\sf\small n = 2m -1 \\$

Substitute this value in equation (i) ,

$\longrightarrow \sf\small \dfrac{2a+ (2m-1-1)d}{2a' +(2m-1-1)d'}=\dfrac{7(2m-1)+1}{4(2m-1) +27}\\$

Simplify,

$\longrightarrow\sf\small \dfrac{ 2a + (2m-2)d}{2a' +(2m-2)d'}=\dfrac{14m-7+1}{8m-4+27}\\$

$\longrightarrow\sf\small \dfrac{2[a + (m-1)d]}{2[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\$

$\longrightarrow\sf\small \dfrac{[a + (m-1)d]}{[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\$

From equation (ii) ,

$\longrightarrow\sf\small \underline{\underline{\blue{ \dfrac{T_{m_1}}{T_{m_2}}=\dfrac{ 14m-6}{8m+23}}}}\\$

$\rule{200}4$

8 0

3 years ago

Find the Tangent of Angle C Find the tangent of ZC. C 8 A 15 17

iren [92.7K]

Answer:

15/8

Step-by-step explanation:

tanC = opposite/adjacent = 15/8

4 0

2 years ago

How do I solve this?

SashulF [63]

Answer: A symbol

Step-by-step explanation: Nothing in that picture gives a value for x

7 0

3 years ago