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4 years ago

10

Dr.Baker and 2 other dentists work in the same office.In one day,the three dentists saw a total 51 patients.Suppose each dentist

saw the same number of patients .How many patients dud each dentist see?

1 answer:

Grace [21]4 years ago

8 0

Well, if the 3 dentists saw 51 patients and they saw a somewhat equal amount, you would have to do 51 divided by 3 and round to the nearest first.

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3 years ago

Yuri thinks that 3/4 is a root of the following function.

sineoko [7]

Given:

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$q(x)=6x^3+19x^2-15x-28$

Yuri thinks that $\dfrac{3}{4}$ is a root of the given function.

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Why $\dfrac{3}{4}$ cannot be a root?

Solution:

We have,

$q(x)=6x^3+19x^2-15x-28$

If $\dfrac{3}{4}$ is a root, then the value of the function at $\dfrac{3}{4}$ is 0.

Putting $x=\dfrac{3}{4}$ in the given function, we get

$q(\dfrac{3}{4})=6(\dfrac{3}{4})^3+19(\dfrac{3}{4})^2-15(\dfrac{3}{4})-28$

$q(\dfrac{3}{4})=6(\dfrac{27}{64})+19(\dfrac{9}{16})-\dfrac{45}{4}-28$

$q(\dfrac{3}{4})=3(\dfrac{27}{32})+\dfrac{171}{16}-\dfrac{45}{4}-28$

$q(\dfrac{3}{4})=\dfrac{81}{32}+\dfrac{171}{16}-\dfrac{45}{4}-28$

Taking LCM, we get

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$q(\dfrac{3}{4})=\dfrac{-833}{32}\neq 0$

Since the value of the function at $\dfrac{3}{4}$ is not equal to 0, therefore, $\dfrac{3}{4}$ is not a root of the given function.

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3 years ago