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Nitella [24]
3 years ago
14

For which value of c will the equation 2x-5=2x-c have an infinite number of solutions?

Mathematics
1 answer:
Vesnalui [34]3 years ago
5 0
There will be infinite answers for x if the equation if the same on both sides, so c=5
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A cyclist travels 5.8 km in 30 minutes. <br> What is his average speed in km/h?
Mila [183]

Answer:

11.6km/hr

Step-by-step explanation:

You need to double 30 minutes to get an hour, so do the same with the distance.

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6 0
2 years ago
What’s is the answer
kvasek [131]

When you are multiplying an exponent directly into a number/variable with an exponent, you multiply the exponents together.

For example:

(x^{2} )^{3} = x^6

(x^{3} )^5=x^{15}


When you are multiplying a variable with an exponent by another variable with an exponent, you add the exponents together.

For example:

(x^{2} )(x^{3})=x^{5}

(x^{1} )(x^{2})=x^{3}


(\frac{(x^{-3})(y^{2})}{(x^{4})(y^{6})} )^{3}=\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}

You multiply 3 into each exponent in the numerator and the denominator

\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}= \frac{y^{6}}{(x^{9})(x^{12})(y^{18})}

When you have a negative exponent, you move it to the other side of the fraction to make the exponent positive.

\frac{y^{6}}{(x^{21})(y^{18})} = \frac{1}{(x^{21})(y^{12})}


When you have something like this:

\frac{x^{2}}{x^5}

You subtract the exponents together, so:

\frac{x^2}{x^5} = x^{2-5} = x^{-3} = \frac{1}{x^3}


Your answer is the second option

3 0
3 years ago
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Answer:

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