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3 years ago

What is the vertex of the graph of the function below?

Mathematics

1 answer:

Dafna1 [17]3 years ago

5 0

The answer is C (-3,-4)

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Find the area and perimeter of the rectangle Length = 5cm Breadth =2cm

avanturin [10]

Answer:

perimeter: 14

Area: 10

Step-by-step explanation:

PERIMETER

2x(L+B) = 2x(5+2) = 2x(7) = 14

AREA

LxB = 5x2 = 10

PLEASE MARK ME BRAINLIEST

5 0

3 years ago

Can somebody please give me a step by step explanation. I have the answer I just don’t know how to show work!!!!!!! Please!!!

krek1111 [17]

Answer:

5391.76

Step-by-step explanation:

We have a rectangle and 2 semicircles

First find the area of the rectangle

A = l*w

A = 44*88= 3872

Then we know that 2 semicircles = 1 circle

The area of a circle is

A = pi r^2

The diameter is 44 so the radius is 22

A = pi * 22^2

A = pi484

A =1519.76

Add the areas together

1519.76+3872

5391.76

7 0

4 years ago

Pls I need help I'm soooooooooooo confused on my question also please no links because I can't get to them!

saul85 [17]

Answer:

divide the least common denominator by the original denominator

6 0

3 years ago

Read 2 more answers

Evaluate. 1. for x=-3 and y=2

Sveta_85 [38]

<h3>Given:</h3>

X = - 3

Y = 2

2^ x^ y^

=2^ (-3)^ (2)^

= (-6)^(2)

3 0

2 years ago

100 POINTS TO RIGHT AWNSER!!!!

makvit [3.9K]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3000586

——————————

The answer is option D) r < 5 or r > – 1.

I'm going to graph each inequality below on a number line.

A) r > 5 or r > – 1.

$\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r>5~~or~~r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}$

The result is found just by joining those two intervals together. Actually that compound inequality only implies

r > – 1

which does not represent all real numbers.

—————

B) r > 5 or r < – 1.

$\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r5~~or~~r$

Numbers between – 1 and 5 (including them) are not included in the union, so you don't have all real numbers represented there either.

—————

C) r < 5 or r < – 1.

$\large\begin{array}{cl} \mathsf{r$

Numbers that are greater or equal to 5 are not in the union. So it does not represent all real numbers.

—————

D) r < 5 or r > – 1.

$\large\begin{array}{cl} \mathsf{r-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r-1}&\qquad\mathsf{\overset{~~~**************************~~~}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}$

Now all real numbers are included in the union. So this is the right choice.

Answer: option D) r < 5 or r > – 1.

I hope this helps. =)

6 0

3 years ago