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VashaNatasha [74]
2 years ago
15

Use compatible numbers to estimate the quotient. 784/8

Mathematics
1 answer:
Furkat [3]2 years ago
5 0
   90   -720 /8
8 - 64/8

784/8 = 98
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6 pints of a 20 percent solution of alcohol in water are mixed with 4 pints of a 10 percent alcohol in water solution. The perce
andreev551 [17]
Total Volume would be 6 + 4 = 10 pints
Let's calculate the volume of alcohol in final solution:
20% * 6 = 1.2 pints
10% * 4 = .4 pints
Total = 1.6 pints of alcohol in 10
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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.

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We have the following equation:

r= \frac{5}{3+2sin(\theta)}

If we graph this equation we realize that in fact this is an ellipse with major axis matching the y-axis. So we can recognize these characteristics:

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The midpoint C<span> of the line segment joining the foci is called the </span>center<span> of the ellipse. So in this exercise this point is as follows:
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C(0, -2)

2. Length of major axis:

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6 \ units

3. Length of minor axis:

The line perpendicular to the foci through the center is called the minor axis. So in the figure if you go from -2, at the x-coordinate, and walk through this minor axis to the coordinate 2, the distance you run is the length of the minor axis, that is:

4 \ units

4. Foci:

Let's find c as follows:

c=\sqrt{a^{2}-b^{2}}=\sqrt{3^{2}-2^{2}}=\sqrt{5}

Then the foci are:

f_{1}=(0, \sqrt{5}-2)

f_{2}=(0, -\sqrt{5}-2)

8 0
2 years ago
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