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Mars2501 [29]
2 years ago
7

Earth is approximately 9.3 × 107 miles from the sun. Saturn is approximately 8.87 × 108 miles from the sun. About how much farth

er is Saturn from the sun than Earth is?
Answers-
A)7.94 × 10 7 miles

B) 7.94 × 10 8 miles

C) 4.3 × 10 7 miles

D) 4.3 × 10 8 miles
Mathematics
1 answer:
7nadin3 [17]2 years ago
7 0

Answer:

B) 7.94 × 10 8 miles

Step-by-step explanation:

Given

Distance between Earth and sun=E=9.3* 〖10〗^7  Miles

Distance Between Saturn and Sun=S=8.87* 〖10〗^8  Miles  

In order to find the distance between Earth and Saturn, we have to subtract the distance between earth and sun from the distance between Saturn and sun. So

Distance between Earth and Saturn=S-E

=(8.87* 10^8 )-( 9.3* 10^7 )

As the distances are in scientific notation, to perform subtraction, the powers of 10 in both quantities have to be same, so

=(8.87*10* 10^7 )-(9.3* 10^7 )

=(88.7* 10^7 )-(9.3* 10^7)

=79.4* 10^7  Miles

It can also be written as:

Distance between Earth and Saturn=7.94* 10^8 Miles  ..

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If eat 1 apple and then i eat another how many apples have i eaten in total
stealth61 [152]

2 apples. Because you ate one already and you eat another one that equals 2

8 0
3 years ago
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You buy 168 pears. There are 28 pears in each bag. How many pears do you buy?
WINSTONCH [101]
To find the answer to your question you would have to divided 168 and 28
And when you do that you will get 6 so the answer is 6 bags
4 0
2 years ago
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Two solutions to y'' – 2y' – 35y = 0 are yı = e, Y2 = e -5t a) Find the Wronskian. W = 0 Preview b) Find the solution satisfying
pashok25 [27]

Answer:

a.w(t)=-12e^{2t}

b.y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}

Step-by-step explanation:

We have a differential equation

y''-2 y'-35 y=0

Auxillary equation

(D^2-2D-35)=0

By factorization method we are  finding the solution

D^2-7D+5D-35=0

(D-7)(D+5)=0

Substitute each factor equal to zero

D-7=0  and D+5=0

D=7  and D=-5

Therefore ,

General solution is

y(x)=C_1e^{7t}+C_2e^{-5t}

Let y_1=e^{7t} \;and \;y_2=e^{-5t}

We have to find Wronskian

w(t)=\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}

Substitute values then we get

w(t)=\begin{vmatrix}e^{7t}&e^{-5t}\\7e^{7t}&-5e^{-5t}\end{vmatrix}

w(t)=-5e^{7t}\cdot e^{-5t}-7e^{7t}\cdot e^{-5t}=-5e^{7t-5t}-7e^[7t-5t}

w(t)=-5e^{2t}-7e^{2t}=-12e^{2t}

a.w(t)=-12e^{2t}

We are given that y(0)=-7 and y'(0)=23

Substitute the value in general solution the we get

y(0)=C_1+C_2

C_1+C_2=-7....(equation I)

y'(t)=7C_1e^{7t}-5C_2e^{-5t}

y'(0)=7C_1-5C_2

7C_1-5C_2=23......(equation II)

Equation I is multiply by 5 then we subtract equation II from equation I

Using elimination method we eliminateC_1

Then we get C_2=-\frac{5}{2}

Substitute the value of C_2 in  I equation then we get

C_1-\frac{5}{2}=-7

C_1=-7+\frac{5}{2}=\frac{-14+5}{2}=-\frac{9}{2}

Hence, the general solution is

b.y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}

7 0
2 years ago
(3.1 x 10^5) (.00000432)​
Ber [7]

Answer:

13.392

Step-by-step explanation:

(3.1×1000000)(.00000432)

7 0
3 years ago
what expressions represent the phrase "twice the sum of a number and 5"? A)2×+5, B) 2 (5)+×, C) 2 (×+5), D) (×+5)^2
Ivanshal [37]
The answer will be c, 2(x+5)
6 0
3 years ago
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