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juin [17]
3 years ago
14

The distance of planet Mercury from the Sun is approximately 5.8 ⋅ 10^7 kilometers, and the distance of Earth from the Sun is 1.

5 ⋅ 10^8 kilometers. About how many more kilometers is the distance of Earth from the Sun than the distance of Mercury from the Sun?
Choices are
A:4.3 ⋅ 107 kilometers 

B:9.2 ⋅ 107 kilometers 

C:9.2 ⋅ 108 kilometers 

D:5.7 ⋅ 109 kilometers
Mathematics
1 answer:
ohaa [14]3 years ago
4 0
1.5X10^8-5.8X10^7

1.5X10^8-0.58X10^8

(1.5-0.58)(10^8)

(0.92)(10^8)

0.92X10^8

9.2X10^7
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13 ft :10 ft 18 ft Find the area of the trapezoid. [? ] square feet Hint: The formula for the area of a trapezoid is: (b17b2). h
Liono4ka [1.6K]
The area is 29473
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3 0
3 years ago
Can someone check my answers to see if they are correct please? They are on the 2nd page. Thank you!
soldier1979 [14.2K]

Answer:

Yes, all your answers are correct.

They are correctly matched with their reasons.

3 0
3 years ago
Help with 21 would be much appreciated.
umka2103 [35]

Answer:

\large\boxed{x=2\sqrt{21}\ and\ y=4\sqrt3}

Step-by-step explanation:

Look at the picture.

ΔADC and ΔCDB are similar. Therefore the sides are in proportion:

\dfrac{AD}{CD}=\dfrac{CD}{DB}

We have

AD=14-8=6\\CD=y\\DB=8

Substitute:

\dfrac{6}{y}=\dfrac{y}{8}         <em>cross multiply</em>

y^2=(6)(8)

y^2=48\to y=\sqrt{48}\\\\y=\sqrt{16\cdot3}\\\\y=\sqrt{16}\ cdot\sqrt3\\\\\boxed{y=4\sqrt3}

For x use the Pythagorean theorem:

x^2=6^2+(4\sqrt3)^2\\\\x^2=36+48\\\\x^2=84\to x=\sqrt{84}\\\\x=\sqrt{4\cdot21}\\\\x=\sqrt4\cdot\sqrt{21}\\\\\boxed{x=2\sqrt{21}}

3 0
3 years ago
Can someone help me asap!<br><br> ​
lilavasa [31]

Answer:

B negative

Step-by-step explanation:

use real numbers?

n is +9

m is -11

m/n * (-m*n^2)

-11/9 * (11*81)

-11/1 * (11*9)

-11 * 99

-1089

6 0
2 years ago
HELPPPPP PLEASEEEE!!!
melamori03 [73]

Answer:

250 MAYBE

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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