All categories

2 years ago

Identify the area of the figure rounded to the nearest tenth. HELP?!! I don't understand!! Please help and show your work!!

Mathematics

1 answer:

Alex73 [517]2 years ago

8 0

Answer:

$67.5\ cm^{2}$

Step-by-step explanation:

we know that

The area of the figure is equal to the area of rectangle (figure 1) plus the area of trapezoid (figure 2)

see the attached figure to better understand the problem

The area of the rectangle is

$A=15*3=45\ cm^{2}$

The area of the trapezoid is

$A=\frac{1}{2}[(15-9)+3)](8-3)$

$A=\frac{1}{2}[6+3)](5)=22.5\ cm^{2}$

The area of the figure is

$45\ cm^{2}+22.5\ cm^{2}=67.5\ cm^{2}$

You might be interested in

Jun used 56 cm of ribbon to go around the edge of a rectangular lid. The lid was 13 cm long. What is the width of the lid?

maksim [4K]

Answer:

The width of the rectangular lid is 15 cm.

Step-by-step explanation:1. Let's review the information given to us to answer the question correctly:

Length of the ribbon used by Jun = 56 cmLength of the rectangular lid = 13 cm

2. What is the width of the lid? The ribbon used by Jun is equivalent to the perimeter of the rectangular lid, therefore:Perimeter = 2 * Length + 2 * WidthReplacing with the values we have:56 = 2 * 13 + 2 * Width56 = 26 + 2Width2Width = 56 - 262Width = 30Width = 30/2Width = 15

5 0

2 years ago

Read 2 more answers

The sum of 3 consecutive integers is 90. Find the largest integer if x is the smallest integer.

____ [38]

Answer:

Step-by-step explanation:

x, x+1, x+2

x + x + 1 +x + 2 = 90

3x + 3 = 90

-3 -3

3x = 87

/3 /3

x = 29

29 + 2 = 31

3 0

3 years ago

A florist received 30 dozen roses. She order 20 dozen. What percent did she receive in excess

KonstantinChe [14]

Extra 10, 10 is half of what she ordered so she received an extra 50%

8 0

3 years ago

Read 2 more answers

What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?

scoray [572]

Answer:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$
Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
$\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify given limit.

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}$

Step 2: Find Limit

Let's start out by directly evaluating the limit:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}$

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

[Limit] Apply Limit Rule [L' Hopital's Rule]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}$
[Limit] Differentiate [Derivative Rules and Properties]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$