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sashaice [31]
2 years ago
14

Help pleasssere!!!!!!!!!!!

Mathematics
1 answer:
kati45 [8]2 years ago
8 0

Step-by-step explanation:

I hope it helps...I just put it in point slope form

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The earth has a mass of approximately 6\cdot 10^{24}6⋅10 24 6, dot, 10, start superscript, 24, end superscript kilograms (\text{
Alex_Xolod [135]

Answer:

0.02

Step-by-step explanation:

The volume of the earth's oceans is approximately 1.34\cdot 10^{9}1.34⋅10

9

1, point, 34, dot, 10, start superscript, 9, end superscript cubic kilometers (\text{km}^3)(km

3

)left parenthesis, start text, k, m, end text, cubed, right parenthesis, and ocean water has a mass of about 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\text{km}^3}1.03⋅10

12

 

km

3

kg

​

1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start text, k, m, end text, cubed, end fraction .

To simplify, we will use the product of powers property of exponents that says that x^a\cdot x^b = x^{a+b}x

a

⋅x

b

=x

a+b

x, start superscript, a, end superscript, dot, x, start superscript, b, end superscript, equals, x, start superscript, a, plus, b, end superscript.

\qquad 1.34\cdot 10^{9}\,\cancel{\text{km}^3} \cdot 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\cancel{\text{km}^3}} = 1.3802 \cdot 10^{21}\,\text{kg}1.34⋅10

9

 

km

3

⋅1.03⋅10

12

 

km

3

kg

​

=1.3802⋅10

21

kg1, point, 34, dot, 10, start superscript, 9, end superscript, start cancel, start text, k, m, end text, cubed, end cancel, dot, 1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start cancel, start text, k, m, end text, cubed, end cancel, end fraction, equals, 1, point, 3802, dot, 10, start superscript, 21, end superscript, start text, k, g, end text

Hint #2

Next we want to know what portion of the earth's mass this represents. We have:

\qquad \begin{aligned} \dfrac{\text{mass of the oceans}}{\text{total mass of the earth}} &= \dfrac{1.3802 \cdot 10^{21}\,\text{kg}}{6\cdot 10^{24}\,\text{kg}} \\\\ &= \dfrac{1.3802}{6\cdot 10^{3}} \\\\ &= \dfrac{1.3802}{6000} \\\\ &= 0.0002300\overline{3} \end{aligned}

total mass of the earth

mass of the oceans

​

​

 

=

6⋅10

24

kg

1.3802⋅10

21

kg

​

=

6⋅10

3

1.3802

​

=

6000

1.3802

​

=0.0002300

3

​

To convert this to a percent, we multiply by 100100100, so the oceans represent 0.02300\overline{3}\%0.02300

3

%0, point, 02300, start overline, 3, end overline, percent of the earth's total mass, according to these figures.

Hint #3

To the nearest hundredth of a percent, 0.020.020, point, 02 percent of the earth's mass is from oceans.

6 0
2 years ago
Please could someone help me with these questions?? brainliest for quickest!
lara [203]

Answer:

523 -61= 462

456-187=269

Step-by-step explanation:

subtract the last numbers first 3-1 which is 2 then 2-6 which is impossible so you borrow from 5 and add 10 to 2 which become 12-6 which is 6 the the first number has only 4 left because you borrowed from it which is 4 so 523-61= 462.

subtract the last number 6-7 it not possible because 6 is less than 7 so borrow from 5 then add 10 to 6 which become 16-7=9 then the second number you have 4-8 since you borrowed from 5, 4-8 is also not possible so borrow from 4 which become 14-8 which is 6 and the first number which is now 3- 1 which is 2 so all the result become 269

7 0
2 years ago
Over 6 days jim jogged 6.5 miles 5 miles 3 miles 2 miles 3.5 miles and 4 miles what is the mean distance that jim jogged each da
soldi70 [24.7K]
In order to identify the mean distance jogged by Jim for 6 days, we add up all of the distances and divide the sum by 6. That is,
                           sum = 6.5 + 5 + 3 + 2 + 3.5 + 4 miles 
                            sum = 24 miles
Dividing the sum by 6 will give us an answer of 4 miles. Thus, Jim jogged an average of 4 miles for 6 days. 
5 0
3 years ago
Read 2 more answers
On your own paper, plot the Points E(−2,2) and F(8,10). Use the vertical and horizontal grid lines to draw a right triangle with
lakkis [162]

Answer:

sssssss

Step-by-step explanation:

sssssssssssssssssssssssssss

6 0
2 years ago
P is the point on the line 2x+y-10=0 such that the length of OP, the line segment from the origin O to P, is a minimum. Find the
nirvana33 [79]
The minimum distance is the perpendicular distance. So establish the distance from the origin to the line using the distance formula.
The distance here is: <span><span>d2</span>=(x−0<span>)^2</span>+(y−0<span>)^2
</span>                                      =<span>x^2</span>+<span>y^2
</span></span>
To minimize this function d^2 subject to the constraint, <span>2x+y−10=0
</span>If we substitute, the y-values the distance function can take will be related to the x-values by the line:<span>y=10−2x 
</span>You can substitute this in for y in the distance function and take the derivative:
<span>d=sqrt [<span><span><span>x2</span>+(10−2x<span>)^2]
</span></span></span></span>
d′=1/2 (5x2−40x+100)^(−1/2)   (10x−40)<span>
</span>Setting the derivative to zero to find optimal x,
<span><span>d′</span>=0→10x−40=0→x=4
</span>
This will be the x-value on the line such that the distance between the origin and line will be EITHER a maximum or minimum (technically, it should be checked afterward).
For x = 4, the corresponding y-value is found from the equation of the line (since we need the corresponding y-value on the line for this x-value).
 
Then y = 10 - 2(4) = 2.
 So the point, P, is (4,2).
8 0
3 years ago
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