Explain how to estimate the quotient using compatiblenumbers.plz help fast

Mathematics

1 answer:

Nadya [2.5K]3 years ago

3 0

<h2><u>PLEASE MARK BRAINLIEST!</u></h2>

Answer:

To estimate the quotient, make the fractions "like" fractions. To do this, you need to make the fractions have the same denominator, and turn them into improper fractions.

Step-by-step explanation:

$27\frac{2}{3} = 27 \frac{20}{30}$

$3\frac{9}{10} = 3\frac{27}{30}$

Now the denominators are the same --> they are "like" fractions.

$27\frac{20}{30} + 3\frac{27}{30} = ?$

$27\frac{20}{30} + 3\frac{27}{30} = (27 + 3) + (\frac{20}{30} + \frac{27}{30})$

$27\frac{20}{30} + 3\frac{27}{30} = (30) + (\frac{47}{30})$

$27\frac{20}{30} + 3\frac{27}{30} = (30) + (1 \frac{17}{30})$

$27\frac{20}{30} + 3\frac{27}{30} = 31 \frac{17}{30}$

Your answer is $31\frac{17}{30}$ .

I hope this helps!

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olga_2 [115]

1. "the graph has the same zeros" : so let a be the "triple" root of the cubic polynomial function.

2. So f(x)= $(x-a)^{3}$

3. Don't forget that the expression might have a coefficient b as well, and still maintain the conditions:

$f(x)=b(x-a)^{3}$

4. Now, f(0)=-5 so $-5=f(0)=b(0-a)^{3}=b(-a) ^{3}=-ba ^{3}$

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$5=ba ^{3}$

$b= \frac{5}{ a^{3} }$

5. the function is $f(x)=\frac{5}{ a^{3} }(x-a)^{3}$ where a can be any real number except 0