All categories

2 years ago

Identify the domain and range

Mathematics

1 answer:

Olenka [21]2 years ago

6 0

Answer:

The domain is (-4, -1].

The range is (-2, 2].

Step-by-step explanation:

The curved parentheses means that the value is not included in the domain/range and the squared one means the value is included.

You might be interested in

When 250 students were interviewed, 96 of the students thought USC was the best school in the nation. The probability that one o

melomori [17]

Answer:

240/96.

Step-by-step explanation:

250-96 =240 so that's 240/96

3 0

2 years ago

When Tomas simplified the expression -2.6 + (-5.4), he got 2.8. what mistake did Tomas likely make?

vladimir2022 [97]

Tomas used incorrectly the rule of signs, the expression should be simplified as follows:

-2.6 + (-5.4)

-2.6 -5.4 = 8

<h3></h3><h3>What mistake did Tomas likely make?</h3>

Here Tomas wants to perform an addition between two numbers:

-2.6 + (-5.4)

And the outcome that Tomas gets is:

-2.6 + (-5.4) = 2.8

Here his mistake seems to bee that he thought the second number was a positive number, and he solved the operation:

-2.6 + 5.4

So, he used wrong the rule of signs.

Remember that the rule of signs says that:

(-)*(+) = (+)*(-) = (-)

Using that, we can rewrite the original expression:

-2.6 + (-5.4)

to:

-2.6 - 5.4

Solving that, we get:

-2.6 - 5.4 = -8

Which is the outcome that Tomas would have gotten if he had used correctly the rule of signs.

If you want to learn more about the rule of signs:

brainly.com/question/13333620

#SPJ1

5 0

1 year ago

Evaluate the following limit:

Makovka662 [10]

If we evaluate the function at infinity, we can immediately see that:

$\large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle L = \lim_{x \to \infty}{\frac{(x^2 + 1)^2 - 3x^2 + 3}{x^3 - 5}} = \frac{\infty}{\infty}} \end{gathered}$}$

Therefore, we must perform an algebraic manipulation in order to get rid of the indeterminacy.

We can solve this limit in two ways.

By comparison of infinities:

We first expand the binomial squared, so we get

$\large\displaystyle\text{$\begin{gathered}\sf \displaystyle L = \lim_{x \to \infty}{\frac{x^4 - x^2 + 4}{x^3 - 5}} = \infty \end{gathered}$}$

Note that in the numerator we get x⁴ while in the denominator we get x³ as the highest degree terms. Therefore, the degree of the numerator is greater and the limit will be \infty. Recall that when the degree of the numerator is greater, then the limit is \infty if the terms of greater degree have the same sign.

Dividing numerator and denominator by the term of highest degree:

$\large\displaystyle\text{$\begin{gathered}\sf L = \lim_{x \to \infty}\frac{x^{4}-x^{2} +4 }{x^{3}-5 } \end{gathered}$}\\$

$\ \ = \lim_{x \to \infty\frac{\frac{x^{4} }{x^{4} }-\frac{x^{2} }{x^{4}}+\frac{4}{x^{4} } }{\frac{x^{3} }{x^{4}}-\frac{5}{x^{4}} } }$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{=\lim_{x \to \infty}\frac{1-\frac{1}{x^{2} } +\frac{4}{x^{4} } }{\frac{1}{x}-\frac{5}{x^{4} } } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{0}=\infty } \end{gathered}$}$

Note that, in general, 1/0 is an indeterminate form. However, we are computing a limit when x →∞, and both the numerator and denominator are positive as x grows, so we can conclude that the limit will be ∞.

5 0

1 year ago

"the quotient of a number and 4"

ladessa [460]

Answer:

a number divided by 4

Step-by-step explanation:

a number = n (it's a variable used in equations.)

so n over 4, or n divided by 4

3 0

2 years ago

Plz guys help i need asap<br> its part b im stuck

aleksklad [387]

Answer:

a.2nd quarter with 9 goals

b. 4.8 goals

c. 4 goals

Step-by-step explanation:

a. The mode is defined as the most appearing data point or the data point with the highest frequency..

From our data(for away goals):

1st quarter-2
2nd quarter-9
3rd quarter-7
4th quarter-4

Hence, the 2nd quarter has the mode for away goals with 9 goals.

b. Mean is defined as the average of a set of data points.

#We calculate the totals goals per quarter, sum over all quarters then divide by the number of games, 10:

$\bar x=\frac{1}{n}\sum{x_i}\\1^{st }_g=Away+Home=5+2=7\\\\2^{nd}_g=Away+Home=4+9=13\\\\3^{rd}_g=Away+Home=8+7=15\\\\4^{th}_g=Away+Home=9+4=13\\\\\bar x=\frac{1}{n}\sum{x_i}=\frac{1}{10}(7+13+15+13)=4.8$

Hence, the mean number of goals per quarter is 4.8 goals

c. To find the number of more home goals than away goals, we subtract from their summations as:

$g_m=\sum{g_h}-\sum{g_a}\\\\=(5+4+8+9)-(2+9+7+4)\\\\=26-22\\\\=4$

Hence, there are 4 more home goals than away goals.

4 0

2 years ago