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Zarrin [17]
3 years ago
14

Find the perimeter of the image below

Mathematics
1 answer:
BabaBlast [244]3 years ago
6 0

Answer:

We need to solve the each side as though it is the hypotenuse:

AB: 6^2 + 5^2 = 61 => 7.81

BC: 8^2 + 5^2 = 89 => 9.43

CD: 4^2 + 3^2 = 25 => 5

DE: 4^2 + 2^2 = 20 => 4.47

EA: 2^2 + 5^2 = 29 => 5.39

Total 32.10 Units

Step-by-step explanation:

Please mark brainliest and have a great day!

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1 - 3/10 Write the answer as a fraction​
IgorLugansk [536]

Answer:

7/10

Step-by-step explanation:

Get a common denominator

1 - 3/10

10/10 - 3/10

7/10

6 0
3 years ago
Read 2 more answers
Y=-5over4x+7 what is the slope pf a line parallel to this line? What is the slope of a line perpendicular?
quester [9]
You question isn't clear, but I assume that you are asking for the slop of a line parallel to y=-(5/4)(x)+7.
The line parallel would have a slop of -5/7 and the line perpendicular would have a slop of 5/7. 
The difference between the two answers is the sign of the two slopes, one is positive the other is negative

4 0
3 years ago
Please please please help
Elina [12.6K]
Median of (72 82 92 93 94 97 98 102)
Median: 93.5
MAD : 7.125

Median of ( 53 59 64 65 65 66 67 69)
Median : 65
MAD : 3.75
4 0
2 years ago
I need help on finals plsssssss
SVEN [57.7K]

Answer:

Our rate is, 10:125

Let's make it a unit rate!

Divide by 10 on both sides:

1:12.5

Step-by-step explanation:

She makes 12.50 dollars per hour.

3 0
3 years ago
Given that f(x) = x + 3<br> a) Find f(2)
Nataly_w [17]

{ \qquad\qquad\huge\underline{{\sf Answer}}}

Here we go ~

In the above question, it is given that :

\qquad \sf  \boxed{ \sf f(x) =  \frac{x + 3}{2} }

A.) Find f(2) :

\qquad \sf  \dashrightarrow \: f(2) =  \dfrac{2 + 3}{2}

\qquad \sf  \dashrightarrow \: f(2) =  \dfrac{5}{2}

or

\qquad \sf  \dashrightarrow \: f(2) = 0.5

B.) Find { \sf {f}^{-1}(x) } :

\qquad \sf  \dashrightarrow \: let \: y = f (x)

so, we can write it as :

\qquad \sf  \dashrightarrow \: y =  \dfrac{x + 3}{2}

\qquad \sf  \dashrightarrow \: 2y = x + 3

\qquad \sf  \dashrightarrow \: x = 2y - 3

Now, put x = { \sf {f}^{-1}(x) }, and y = x and we will get our required inverse function ~

\qquad \sf  \dashrightarrow \: f {}^{ - 1}(x) = 2x- 3

C.) Find { \sf {f}^{-1}(12) } :

\qquad \sf  \dashrightarrow \: f {}^{ - 1}(12) = 2(12)- 3

\qquad \sf  \dashrightarrow \: f {}^{ - 1}(12) = 36- 3

\qquad \sf  \dashrightarrow \: f {}^{ - 1}(12) = 33

7 0
1 year ago
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