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2 years ago

Find the perimeter of this shape please help I will award brainlisest

Mathematics

2 answers:

viktelen [127]2 years ago

8 0

Answer:

The shape is made up of a square and a rectangle.

So calculating the area of both the figures and then adding them would give us the required result.

Area of square = (a)² = (4)² = 16 sq unit.

Area of rectangle = l×b = 3×2 = 6 sq unit.

Now, the required area = 16 + 6 = 22 sq unit.

stira [4]2 years ago

5 0

Answer:

22 cm

Step-by-step explanation:

You might be interested in

What is the solution to x/3 = -12

Mademuasel [1]

Answer:

x = -36

Step-by-step explanation:

Just multiply both sides of the equation with 3...,which tend to give a solution of,

; x = -12(3)

; x = -36

7 0

3 years ago

nter an equation and an answer to the following question: What is the original list price if the discount is $25 and the discoun

castortr0y [4]

Let us set up an equation, to determine the list price. X represents the original price

.12 * x = 25

In order to solve for X, you must divide .12 from both sides

x = 25/.12

After you divide, you should get the number 208.33

25/.12 = 208.33

So, the original list price is $208.33

7 0

3 years ago

Read 2 more answers

Find the domain of the function y = 3 tan(23x)

solmaris [256]

Answer:

$\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

In other words, the $x$ in $f(x) = 3\, \tan(23\, x)$ could be any real number as long as $x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)$ for all integer $k$ (including negative integers.)

Step-by-step explanation:

The tangent function $y = \tan(x)$ has a real value for real inputs $x$ as long as the input $x \ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

Hence, the domain of the original tangent function is $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

On the other hand, in the function $f(x) = 3\, \tan(23\, x)$ , the input to the tangent function is replaced with $(23\, x)$ .

The transformed tangent function $\tan(23\, x)$ would have a real value as long as its input $(23\, x)$ ensures that $23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

In other words, $\tan(23\, x)$ would have a real value as long as $x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right)$ .

Accordingly, the domain of $f(x) = 3\, \tan(23\, x)$ would be $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

4 0

2 years ago

caleb and emily are standing 100 yards from each other. caleb looks up at a 45 degree angle to see a hot air balloon. Emily look

Sunny_sXe [5.5K]

Check the picture below.

thus then

$\bf \begin{cases}
y=100-a\\
y=a\sqrt{3}
\end{cases}\implies 100-a=a\sqrt{3}\implies 100=a+a\sqrt{3}
\\\\\\
100=a(1+\sqrt{3})\implies \boxed{\cfrac{100}{1+\sqrt{3}}=a}
\\\\\\
\textit{what is the altitude \underline{y} then?}\qquad y=100-a
\\\\\\
y=100-\cfrac{100}{1+\sqrt{3}}\implies y\approx 63.39745962156~yards$

8 0

3 years ago

Read 2 more answers

a toy store marks down every toy by 15% in January. How much does a toy cost during January? Use p for the price in December.

satela [25.4K]

A toy cost 0.85p in January.

Step-by-step explanation:

Given,

Mark down in January = 15%

Let,

Price in December = p

Mark down in January = 15% of price in December

Mark down in January = $\frac{15}{100}*p$

Mark down in January = 0.15p

Price of toy in January = Price in December - Mark down

Price of toy in January = p - 0.15p = 0.85p

A toy cost 0.85p in January.

Keywords: mark down, subtraction

Learn more about subtraction at:

#LearnwithBrainly

7 0

3 years ago