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Sunny_sXe [5.5K]
3 years ago
10

Find the area please!

Mathematics
1 answer:
KiRa [710]3 years ago
3 0

Answer:

you have to divide the shape into a semi-circle and 2 triangles

Step-by-step explanation:

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There is a famous irrational number called Euler's number, symbolized with an e. Like π , its decimal fo rm never ends or repeat
defon

\sqrt{7} < e < \sqrt{8}=2\sqrt{2}

3 0
3 years ago
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What is the solution to the system of equations Use the substitution method to solve the system of equations. Show your work.
marishachu [46]

Answer:

Step-by-step explanation:

Let's work to solve this system of equations:

y = 2x ~~~~~~~~\gray{\text{Equation 1}}y=2x        Equation 1

x + y = 24 ~~~~~~~~\gray{\text{Equation 2}}x+y=24        Equation 2

The tricky thing is that there are two variables, xx and yy. If only we could get rid of one of the variables...

Here's an idea! Equation 11 tells us that \goldD{2x}2x and \goldD yy are equal. So let's plug in \goldD{2x}2x for \goldD yy in Equation 22 to get rid of the yy variable in that equation:

\begin{aligned} x + \goldD y &= 24 &\gray{\text{Equation 2}} \\\\ x + \goldD{2x} &= 24 &\gray{\text{Substitute 2x for y}}\end{aligned}  

x+y

x+2x

​    

=24

=24

​    

Equation 2

Substitute 2x for y

​  

Brilliant! Now we have an equation with just the xx variable that we know how to solve:

x+2x3x 3x3x=24=24=243=8Divide each side by 3

Nice! So we know that xx equals 88. But remember that we are looking for an ordered pair. We need a yy value as well. Let's use the first equation to find yy when xx equals 88:

\begin{aligned} y &= 2\blueD x &\gray{\text{Equation 1}} \\\\ y &= 2(\blueD8) &\gray{\text{Substitute 8 for x}}\\\\ \greenD y &\greenD= \greenD{16}\end{aligned}  

y

y

y

​    

=2x

=2(8)

=16

​    

Equation 1

Substitute 8 for x

​  

Sweet! So the solution to the system of equations is (\blueD8, \greenD{16})(8,16). It's always a good idea to check the solution back in the original equations just to be sure.

Let's check the first equation:

\begin{aligned} y &= 2x \\\\ \greenD{16} &\stackrel?= 2(\blueD{8}) &\gray{\text{Plug in x = 8 and y = 16}}\\\\ 16 &= 16 &\gray{\text{Yes!}}\end{aligned}  

y

16

16

​    

=2x

=

?

2(8)

=16

​    

Plug in x = 8 and y = 16

Yes!

​  

Let's check the second equation:

\begin{aligned} x +y &= 24 \\\\ \blueD{8} + \greenD{16} &\stackrel?= 24 &\gray{\text{Plug in x = 8 and y = 16}}\\\\ 24 &= 24 &\gray{\text{Yes!}}\end{aligned}  

x+y

8+16

24

​    

=24

=

?

24

=24

​    

Plug in x = 8 and y = 16

Yes!

​  

Great! (\blueD8, \greenD{16})(8,16) is indeed a solution. We must not have made any mistakes.

Your turn to solve a system of equations using substitution.

Use substitution to solve the following system of equations.

4x + y = 284x+y=28

y = 3xy=3x

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4 years ago
. Evaluate | –2 | = ?
Basile [38]

Answer:

2

Step-by-step explanation:

6 0
3 years ago
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Prove that root 7 is irrational by the method of contradiction
Alchen [17]

Let assume that \sqrt7 is a rational number. Therefore it can be expressed as a fraction \dfrac{a}{b} wherea,b\in\mathbb{Z} and \text{gcd}(a,b)=1.

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This means that a^2 is divisible by 7, and therefore also a is divisible by 7.

So, a=7k where k\in\mathbb{Z}

(7k)^2=7b^2\\\\49k^2=7b^2\\\\7k^2=b^2

Analogically to a^2=7b^2 ------- b^2 is divisible by 7 and therefore so is b.

But if both numbers a and b are divisible by 7, then \text{gcd}(a,b)=7 which contradicts our earlier assumption that \text{gcd}(a,b)=1.

Therefore \sqrt7 is an irrational number.

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4 years ago
A rectangle is 2/5 in. long and 1/3 in. wide.what is the area
pashok25 [27]
To find the area of a rectangle, simply multiply the length by the width.

2/5 * 1/3 = 2/15

So, the area is 2/15.
6 0
4 years ago
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