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mestny [16]
3 years ago
14

Using Properties to Simplify Expressions

Mathematics
1 answer:
Fantom [35]3 years ago
4 0

Answer:

1) 2x-3 = x -3 + x

2) 2x - 6 = -6 + 2x

3) \frac{3}{5}-\frac{13}{5}=-6.9+4.9

4) 3x+0 = 3x

5)  \frac{5}{3}-x+\frac{1}{3}=-\frac{3}{2}x+2+\frac{1}{2}x

Step-by-step explanation:

We need to match each expression on the left with an equivalent expression on the right

The expressions are equivalent if the have the same results.

1) 2x - 3

Looking at the options the best match is:

x -3 + x

When solve we get:

2x-3

So, 2x-3 = x -3 + x

2) 2x - 6

Looking at the options the best match is:

-6 + 2x

Because according to commutative property: a+b = b+a

So, 2x - 6 = -6 + 2x

3) \frac{3}{5}-\frac{13}{5}

First simplifying the given expression:

\frac{3}{5}-\frac{13}{5}\\=\frac{3-13}{5}\\=\frac{-10}{5}\\=-2

Looking at the options, -6.9+4.9 = -2

So, \frac{3}{5}-\frac{13}{5}=-6.9+4.9

4) 3x + 0

Looking at the options the best match is:

3x

Because, adding 0 in 3x will result in 3x

So, 3x+0 = 3x

5) \frac{5}{3}-x+\frac{1}{3}

First simplifying:

\frac{5}{3}-x+\frac{1}{3}\\=\frac{5-3x+1}{3}\\=\frac{6-3x}{3}\\=  \frac{3(2-x)}{3}\\=2-x

Now, solving the option: -\frac{3}{2}x+2+\frac{1}{2}x

=\frac{-3x+4+1x}{2}\\=\frac{-2x+4}{2}\\=\frac{2(-x+2)}{2}\\=-x+2\\or\\=2-x

So, \frac{5}{3}-x+\frac{1}{3}=-\frac{3}{2}x+2+\frac{1}{2}x

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For problems like this we can use algebra to help us, the farenheight 6 hours ago was:
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Craig has 36 ounces of flour left in one bag and 64 ounces of flour in another bag. Use the Baking Flour Equivalent table to fin
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So 16 = 3.6, 10 = 2.3 so 20 = 4.6 then add 16 + 20 to get 36 so 3.6 + 4.6 = 8.2
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3 0
4 years ago
A spherical balloon currently has a radius of 19cm. If the radius is still growing at a rate of 5cm or minute, at what rate is a
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Answer:

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Step-by-step explanation:

Given:

Radius of the balloon at a certain time (r) = 19 cm

Rate of growth of radius is, \frac{dr}{dt}=5\ cm/min

The rate at which the air is pumped in the balloon can be calculated by finding the rate of increase in the volume of the balloon.

So, first we find the volume of the sphere in terms of 'r'. As the balloon is spherical in shape, the volume of the balloon is equal to the volume of a sphere. Therefore,

Volume of balloon is given as:

V=\frac{4}{3}\pi r^3

Now, rate of increase of volume is obtained by differentiating both sides of the equation with respect to time 't'.

Differentiating both sides with respect to time 't', we get:

\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)\\\\\frac{dV}{dt}=\frac{4\pi}{3}(3r^2)(\frac{dr}{dt})\\\\\frac{dV}{dt}=4\pi r^2(\frac{dr}{dt})

Now, plug in 19 cm for 'r', 5 cm per minute for \frac{dr}{dt} and solve for \frac{dV}{dt}. This gives,

\frac{dV}{dt}=4\pi (19 cm)^2(5\ cm/min)\\\\\frac{dV}{dt}=4\times 3.14\times 361\times 5\ cm^3/min\\\\\frac{dV}{dt}=22670.8\ cm^3/min

Therefore, the rate at which the air is being pumped into the balloon is 22670.8 cm³/min.

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