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3 years ago

What is 0.368 in expanded form

Mathematics

1 answer:

kondor19780726 [428]3 years ago

3 0

The expanded form of 0.368 is

.3 + .06 + .008

Hope this helps. :)

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What is the sum of the polynomials (-x^2+9)+(-3x^2-11x+4)

Marta_Voda [28]

You can just add the terms:

<span>(-x^2+9)+(-3x^2-11x+4) = -4x</span>² - 11x + 13

6 0

3 years ago

Read 2 more answers

1: n/-3 - 8 < 17<br> 2: n/-2 + 11 > 17

cupoosta [38]

Answer:

1: n > -75

2: n < -12

Step-by-step explanation:

1: n/-3 - 8 < 17

n/-3 (- 8 + 8) < 17 + 8

n/-3 < 25

n/-3(-3) < 25(-3)

n < -75

n > -75 (switch symbol when you divide or multiply by a negative number)

2: n/-2 + 11 > 17

n/-2 (+ 11 - 11) > 17 - 11

n/-2 > 6

n/-2(-2) > 6(-2)

n > -12

n < -12 (switch symbol when you divide or multiply by a negative number)

7 0

2 years ago

Which fraction is less than 1 chose all of the correct answer, and explain why you chose and write it down and explain your answ

EleoNora [17]

Answer:

$(3/4)$ and $(2/3)$ .

Step-by-step explanation:

For a fraction of the form $(p / q)$ , the number in front of the division ( $p$ ) is the numerator. The number after the division ( $q$ ) is the denominator.

$\begin{aligned} \frac{p}{q} \quad \genfrac{}{}{0}{}{(\text{numerator})}{(\text{denominator})}\end{aligned}$ .

A positive fraction $(p/q)$ ( $p,\, q > 0$ ) is less than $1$ as long as the numerator $p$ is smaller than the denominator $q$ (that is, $p < q$ .) The reason is that as long as $p,\, q > 0$ :

$\begin{aligned} & \frac{p}{q} < 1 \\ \iff & (q)\, \left(\frac{p}{q}\right) < (q)\, (1) && (\text{given that $q > 0$}) \\ \iff & p < q\end{aligned}$ .

For example, in the fraction $(2/3)$ , $2$ is the numerator while $3$ is the denominator. Both the numerator and the denominator are positive, while the numerator $2\!$ is smaller than the denominator $3\!$ . Therefore, $(2/3)$ would be less than $1$ .