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Ksju [112]
3 years ago
11

Help me!!! When the following expression is written in the simplest form, what is the coefficient of the variable term?

Mathematics
1 answer:
Sindrei [870]3 years ago
5 0

Answer:

I believe the answer is 0

Step-by-step explanation:

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On a given planet, the weight of an object varies directly with the mass of the object. Suppose that an object whose mass is 9 k
nadya68 [22]

Answer:

6 kg

Step-by-step explanation:

Given is weight of an object directly varies with the mass of the object .

Let the constant of proportionality be k

So,

Weight = K×(mass of the object)

For an object , weight = 45 N

                        mass = 9 kg

So , 45 = K×9

      K = 5.

Now, for the other object ,

Weight = 30 N. K = 5 (as obtained above)

So, 30 = 5×(mass of the object)

mass of the object = 6 kg.

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Solve the system of equations:<br> y = 2x - 3<br> y = x2 - 3
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(0, -3), (2, 1)

Step-by-step explanation:

Solve this by solving for the first variable in one of the equations, then substitute the result into the other equation.

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4 years ago
Drop your snap if I can talk to you about a seventh grade math test on circumference and angles?
Dennis_Churaev [7]

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3 years ago
Read 2 more answers
Explain how to get that answer!!
ra1l [238]
We need to simplify \frac{ \sqrt{14x^3} }{ \sqrt{18x} }

First lets factor \sqrt{14x^3}

\sqrt{14x^3} = \sqrt{14}  \sqrt{x^3}
\sqrt{14} =  \sqrt{2} \sqrt{7} by applying the radical rule \sqrt[n]{ab} =  \sqrt[n]{a} \sqrt[n]{b}
\sqrt{x^3} = x^{3/2} By applying the radical rule \sqrt[n]{x^m} = x^{m/n}

So
\sqrt{14x^3} = \sqrt{14}  \sqrt{x^3} = \sqrt{2} \sqrt{7}x^{3/2}

Now let's factor \sqrt{18x}
By applying the radical rule \sqrt[n]{ab} =  \sqrt[n]{a}  \sqrt[n]{b},
\sqrt{18x} =  \sqrt{18} \sqrt{x}
\sqrt{18} =  \sqrt{2} * 3

So \sqrt{18x} = \sqrt{2}*3 \sqrt{x}

So  \frac{ \sqrt{14x^3} }{ \sqrt{18x} } = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}  }

We know that \sqrt[n]{x} = x^{1/n} so \sqrt{x} = x^{1/2}

We now have \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}} = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}}

We know that \frac{x^a}{x^b} = x^{a-b}
So \frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x

We now got \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}&#10;

We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
\frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3} =   \frac{ \sqrt{7}x }{3}


All in All, we get \frac{ \sqrt{14x^3} }{ \sqrt{18x} } =  \frac{ \sqrt{7}x }{3}

Hope this helps! :D


6 0
3 years ago
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