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kipiarov [429]
2 years ago
15

What are the advantages of using a table to find equivalent ratios?

Mathematics
1 answer:
prisoha [69]2 years ago
8 0

The advantage of using a table to determine the equivalent ratio is that, it easier to determine the unknown ratio.

The advantage of using a graph to determine the equivalent ratio is that, it helps to determine the exact value of the unknown ratio.

The use of table to find equivalent ratio makes it easier to determine the unknown ratio. The unknown ratio can easily be determined using interpolation method, this method gives the approximate value of the unknown ratio.

The use of graph to find equivalent ratio helps to determine the exact value of the unknown ratio. This method gives the exact value of the unknown ratio.

The use of table is better because it minimizes error from tracing the values from the graph.

Learn more about equivalent ratio here: brainly.com/question/13513438

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The diagram shows an empty cylindrical container. Ana puts a solid cube of side length 8 cm into the container. She then pours 1
marshall27 [118]

Answer:

<h3><u>Cylinder</u></h3>

\textsf{Volume of a cylinder}=\sf \pi r^2 h \quad\textsf{(where r is the radius and h is the height)}

Given:

  • r = 6 cm
  • h = 18 cm

\begin{aligned}\implies \sf V &= \sf \pi (6)^2(18)\\& = \sf 648 \pi \: cm^3\end{aligned}

<h3><u>Cube</u></h3>

\textsf{Volume of a cube}=\sf x^3\quad\textsf{(where x is the side length)}

Given:

  • x = 8 cm

\begin{aligned}\implies \sf V &= 8^3\\& = \sf 512 \: cm^3\end{aligned}

<h3><u>Volume available to be filled with water</u></h3>

Volume of cylinder - volume of cube

= 684π - 512

= 1532.75204 cm³

    1 litre = 1000 cm³

⇒ 1.5 litres = 1000 × 1.5 = 1500 cm³

As 1500 < 1532.75204, the volume of water poured into the container is smaller than the empty space available in the cylinder.  Therefore, the water will <u>not</u> come over the top of the container.

4 0
2 years ago
I need help solving this 9+18s-7s help me solve it out please
melomori [17]

Answer:

9+11s

Step-by-step explanation:

9+18s-7s

=9+11s

You can't add 9 and 11s because they are different variables

4 0
3 years ago
Read 2 more answers
Find the angle of cd round to the nearest tenth
ololo11 [35]
Answer:
14.1
Step-by-step explanation:

6 0
2 years ago
Reposting this because no body gave me the equation. An equation would be appreciated :)
o-na [289]

bring the like terms together,

\frac{6q}{5}  +  \frac{4q}{5}  \leqslant 28 + 3 + 3 - 3

\frac{10q}{5}  \leqslant 31

2q \leqslant 31

q \leqslant  \frac{31}{2}

3 0
3 years ago
Read 2 more answers
The force of gravity on Mars is different than on Earth. The function of the same situation on Mars would be represented by the
sweet-ann [11.9K]

Answer:

If thrown up with the same speed, the ball will go highest in Mars, and also it would take the ball longest to reach the maximum and as well to return to the ground.

Step-by-step explanation:

Keep in mind that the gravity on Mars; surface is less (about just 38%) of the acceleration of gravity on Earth's surface. Then when we use the kinematic formulas:

v=v_0+a\,*\,t\\y-y_0=v_0\,* t + \frac{1}{2} a\,\,t^2

the acceleration (which by the way is a negative number since acts opposite the initial velocity and displacement when we throw an object up on either planet.

Therefore, throwing the ball straight up makes the time for when the object stops going up and starts coming down (at the maximum height the object gets) the following:

v=v_0+a\,*\,t\\0=v_0-g\,*\,t\\t=\frac{v_0}{t}

When we use this to replace the 't" in the displacement formula, we et:

y-y_0=v_0\,* t + \frac{1}{2} a\,\,t^2\\y-y_0=v_0\,(\frac{v_0}{g} )-\frac{g}{2} \,(\frac{v_0}{g} )^2\\y-y_0=\frac{1}{2} \frac{v_0^2}{g}

This tells us that the smaller the value of "g", the highest the ball will go (g is in the denominator so a small value makes the quotient larger)

And we can also answer the question about time, since given the same initial velocity v_0 , the smaller the value of "g", the larger the value for the time to reach the maximum, and similarly to reach the ground when coming back down, since the acceleration is smaller (will take longer in Mars to cover the same distance)

3 0
3 years ago
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