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maw [93]
2 years ago
12

Nevin made tables of values to approximate the solution to a system of equations. First he found that the x-value of the solutio

n was between one and two, and then he found that it was between one and 1.5 next, he made his table.

Mathematics
2 answers:
prisoha [69]2 years ago
7 0

Answer:

The correct option is B.

Step-by-step explanation:

The given equation are

y=4x-3

y=-5x+9

On solving both the equation, we get

4x-3=-5x+9

4x+5x=9+3

9x=12

9x=\frac{12}{9}=\frac{4}{3}

Put this value in the given equation.

y=4(\frac{4}{3})-3

y=\frac{16}{3}-3

y=\frac{16-9}{3}

y=\frac{7}{3}

The solution of the given system of equation is

(\frac{4}{3},\frac{7}{3})=(1.333,2.333)\approx (1.3,2.3)

The best approximation of the exact solution is (1.3,2.3). Therefore the correct option is B.

Leokris [45]2 years ago
6 0

Answer:

B. (1.3, 2.3)

Step-by-step explanation:

To find approximate solution of the system of equations, we need to equivate the two given equations and solve for x and y.

y = 4x - 3 and y = -5x + 9

4x - 3 = -5x + 9

Isolating the variables and constants, we get

4x + 5x = 9 + 3

9x = 12

Dividing both sides by 9, we get

x = 12 ÷ 9

x = \frac{4}{3}

x = 1.3

Now plug in x =  \frac{4}{3} in the equation y = 4x - 3 to find the value of y.

y = 4( \frac{4}{3})  - 3

y = \frac{16}{3} - 3

y =  \frac{16 - 9}{3}

y =  \frac{7}{3}

y = 2.3

So, x = 1.3 and y = 2.3 is the approximate solution.

Therefore, answer is B. (1.3, 2.3)

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8.1g of sugar is needed for every cake made. How much sugar is needed for 6 cakes?
Lerok [7]

Answer:

48.6

Step-by-step explanation:

If you use 8.1g of sugar for 1 cake then 6 cakes will be 48.6g of sugar

Just do 8.1*6 and you will get 48.6

8 0
2 years ago
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
1 year ago
In a science experiment, Shelly has to weigh a compound in grams. The balance in her school measures in milligrams (mg). Shelly
Oksi-84 [34.3K]
We know, 1 g = 1000 mg
so, 1 mg = 1/1000 g
then, 8,450 mg = 1/1000 * 8,450 = 8.450 g

In short, Your Answer would be: 8.450 Grams

Hope this helps!
5 0
2 years ago
Read 2 more answers
The function D(t) defines a traveler's distance from home, in miles, as a function of time, in hours. D(t) = StartLayout enlarge
nalin [4]

- At 2 hours, the traveler is 725 miles from home.

- At 3 hours, the distance is constant, at 880 miles. --> TRUE.

- The total distance from home after 6 hours is 1,062.5 miles --> TRUE.

Step-by-step explanation:

The function D(t) is defined as follows:

D(t) = 300t+125 for t< 2.5

D(t) = 880 for 2.5 \leq 3.5

D(t) = 75t+612.5 for t\leq 6

Where

t is the time in hours

D(t) is the distance covered, in miles, after t hours

Now let's analyze the different statements:

- The starting distance, at 0 hours, is 300 miles. --> FALSE. In fact, if we substitute t = 0 into the 1st equation, we get

D(0) = (300)(0)+125 = 125

So, the distance at t = 0 is 125 miles.

- At 2 hours, the traveler is 725 miles from home. --> TRUE. In fact, if we substitute t = 2 into the 1st equation,

D(2) = (300)(2)+125 = 725

- At 2.5 hours, the traveler is 875 miles from home. --> FALSE. In fact, for t=2.5 we have to use the 2nd equation, which states that the distance is:

D(t) = 880

So, not 875 miles.

- At 3 hours, the distance is constant, at 880 miles. --> TRUE. This is clearly visible from the 2nd equation: for t between 2.5 and 3.5 (so, in this case), the distance is

D(t) = 880

- The total distance from home after 6 hours is 1,062.5 miles --> TRUE. In fact, if we replace t = 6 into the last equation,

D(6)) = 75(6)+612.5=1062.5

Learn more about functions:

brainly.com/question/3511750

brainly.com/question/8243712

brainly.com/question/8307968

Learn more about  distance:

brainly.com/question/3969582

#LearnwithBrainly

4 0
1 year ago
Read 2 more answers
How much heat is required to raise the temperature of
Elina [12.6K]

Answer:

The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .

Step-by-step explanation:

Given as :

The mass of liquid water = 50 g

The initial temperature = T_1 = 15°c

The final temperature = T_2  = 100°c

The latent heat of vaporization of water = 2260.0 J/g

Let The amount of heat required to raise temperature = Q Joule

Now, From method

Heat = mass × latent heat × change in temperature

Or, Q = m × s × ΔT

or, Q =  m × s × ( T_2 - T_1 )

So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )

Or, Q =  50 g × 2260.0 J/g × 85°c

∴   Q = 9,605,000  joule

Or, Q =  9,605 × 10³ joule

Or, Q = 9605 kilo joule

Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer

4 0
1 year ago
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