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Lelechka [254]
3 years ago
10

Please find the area of the figure!!

Mathematics
1 answer:
Murrr4er [49]3 years ago
4 0

The figure consists of a rectangle and a circle.

first lets find the sides of the triangle

sides of the triangle are 4+4=8 and 3+1=4

therefore area of the rectangle is 8*4=32

now area of semicircle =πr^2/2 = π2^2/2=2π=6.3

we have to add the area of rectangle with the area of semicircle

this is because it the given figure no part of area of circle and rectangle are in common.

therefore total area of given figure = 32+6.3=38.3

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AlexFokin [52]

C

100 divided by 2 (boy and girl) is 50, then 4 goes into 100 25 times so a chance of being a girl in a family of 4 is 25% or .25

5 0
2 years ago
36 jelly donuts is 200 percent of how<br> many jelly donuts?
Art [367]

Answer:

18

Step-by-step explanation:

Equation

2x=36

Solve

x=18

7 0
3 years ago
Read 2 more answers
A teacher is randomly calling on students in a class. If there are 6 girls and 6 boys in the class, what is the theoretical prob
Evgesh-ka [11]

It would be a 50/50 chance that all of them would be girls

6 0
3 years ago
Is the product of 12 and 3/4 more or less than 12?
faust18 [17]
Less, 12 x 3/4 is 9 as you take 12/4 which is 3 times 3, which gives you 9
3 0
3 years ago
Which statement describes if there is an extraneous solution to the equation √x-3 = x-5? A. there are no solutions to the equati
Brrunno [24]
Remember that <span>an extraneous solution of an equation, is the solution that emerges from solving the equation but is not a valid solution.
 
Lets solve our equation to find out what is the extraneous solution:
</span>\sqrt{x-3} =x-5
(\sqrt{x-3})^2 =(x-5)^2
x-3=x^2-10x+25
x^2-11x+28=0
(x-4)(x-7)=0
x-4=0 and x-7=0
x=4 and x=7
<span>
So, the solutions of our equation are </span>x=4 and x=7. Lets replace each solution in our original equation to check if they are valid solutions:
- For x=7
\sqrt{x-3} =x-5
\sqrt{7-3} =7-5
\sqrt{4} =2
2=2
We can conclude that 7 is a valid solution of the equation.

- For x=4
\sqrt{x-3} =x-5
\sqrt{4-3} =4-5
\sqrt{1} =1
1 \neq 1
We can conclude that 4 is not a valid solution of the equation; therefore, 4 is a extraneous solution.

We can conclude that the correct answer is: <span>D. the extraneous solution is x = 4</span>
7 0
3 years ago
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