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aalyn [17]
3 years ago
9

Please answer fast please

Mathematics
1 answer:
kvv77 [185]3 years ago
4 0

Answer:

you can use a website called calculator soup the answer is -9/14 please mark me branlist

Step-by-step explanation:

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All the numbers round to 7 except<br><br><br> 6.61<br><br> 7.89<br><br> 6.58<br><br> 7.37
alisha [4.7K]
7.89 rounds off to 8
6 0
2 years ago
Read 2 more answers
Polygon F has an area of 36 square units. Aimar drew a scaled version of Polygon F and labeled it Polygon G. Polygon G has an ar
Free_Kalibri [48]

Answer:

1/3

Step-by-step explanation:

The area of Polygon GGG is \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction the area of Polygon FFF.

Each side of Polygon FFF was multiplied by a certain value, known as the scale factor , to result in an area that is \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction the area of Polygon FFF.

[Show me an example of how scale factor affects area]

\dfrac1{10}  

start fraction, 1, divided by, 10, end fraction

 

 

\begin{aligned} A &= \left(l\times\dfrac1{10}\right)\times\left(w\times\dfrac1{10}\right) \\ \\ A&= l\times w\times\dfrac1{10}\times\dfrac1{10} \\ \\ A&= lw \times \left(\dfrac1{10}\right)^2\end{aligned}  

 

 

 

 

 

 

 

 

\dfrac1{10}  

start fraction, 1, divided by, 10, end fraction\left(\dfrac1{10}\right)^2  

 

left parenthesis, start fraction, 1, divided by, 10, end fraction, right parenthesis, start superscript, 2, end superscript

Hint #22 / 3

The area of a polygon created with a scale factor of \dfrac1x  

x

1

​  start fraction, 1, divided by, x, end fraction has \left(\dfrac1{x}\right)^2(  

x

1

​  )  

2

left parenthesis, start fraction, 1, divided by, x, end fraction, right parenthesis, start superscript, 2, end superscript the area of the original polygon:

\left(\text{scale factor}\right)^2=\text{fraction of the area the scale copy has}(scale factor)  

2

=fraction of the area the scale copy hasleft parenthesis, s, c, a, l, e, space, f, a, c, t, o, r, right parenthesis, start superscript, 2, end superscript, equals, f, r, a, c, t, i, o, n, space, o, f, space, t, h, e, space, a, r, e, a, space, t, h, e, space, s, c, a, l, e, space, c, o, p, y, space, h, a, s

The area of Polygon GGG is \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction the area of Polygon FFF. Let's substitute \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction into the equation to find the scale factor.

\left(\dfrac1{?}\right)^2=\dfrac19(  

?

1

​  )  

2

=  

9

1

​  left parenthesis, start fraction, 1, divided by, question mark, end fraction, right parenthesis, start superscript, 2, end superscript, equals, start fraction, 1, divided by, 9, end fraction

The scale factor is \dfrac13  

3

1

​  start fraction, 1, divided by, 3, end fraction.

Hint #33 / 3

Aimar used a scale factor of \dfrac13  

3

1

​  start fraction, 1, divided by, 3, end fraction to go from Polygon FFF to Polygon GGG.

4 0
3 years ago
Read 2 more answers
Find the solutions of this system of equations 10x-5y=-20 7x-5y=-26
OverLord2011 [107]

Answer:

i think $25.00

Step-by-step explanation:


8 0
3 years ago
Which of these relations on{0,1,2,3}are partial orderings? Determine the properties of a partial ordering that the others lack.
omeli [17]

Step-by-step explanation:

A = {0,1,2,3}

a): R = {(0,0),(2,2),(3,3)}

R is antisymmetric, because if the (a,b)∈R, than a=b.

R is not reflexive, because (1,1) ∉ R while 1 ∈ A.

R is transitive, because if the (a,b)∈R and (b, c) ∈ R, than a=b=c and (a,c)=(a,a)∈R.

R is not portable ordering because R is not reflexive.

b): R = {(0,0),(1,1),(2,0),(2,2),(2,3),(3,3)}

R is antisymmetric, because if the (a,b)∈R and if the (b, a) ∈ R, than a=b (since (2,0) ∈ R and (0,2) ∉ R; and (2,3) ∈ R and (3,2) ∉ R )

R is reflexive, because (a,a) ∈ R of every element a ∈ A.

R is transitive , because if the (a,b)∈R and if the ( b , c )∈R . then a = b or b = c ( since there are only two element not of the form ( a , a ) and that pair does not satisfy ( a,b ) ∈ R and ( b , a ) ∈ R ), which implies ( a , c ) = ( b , c ) ∈ R or ( a , c ) = ( a , b ) ∈ R.

R is a partial ordering, because R is reflexive, antisymmetric and transitive.

c): R =  {(0,0),(1,1),(1,2),(2,2),(3,1),(3,3)}

R is reflexive, because (a,a)∈R of every element a ∈ A.

R is antisymmetric, because if the ( a , b )∈R and if the ( b , a )∈R . then a = b ( since ( 1 , 2 )∈R and ( 2 , 1 ) ∉ R; ( 3 , 1 ) ∈ R and ( 1 , 3 ) ∉ R ).  

R is not transitive , because ( 3 , 1 ) ∈ R and ( 1 , 2 )∈R, while ( 3 , 2 ) ∈ R.

R is not a partial ordering. because R is not transitive .

d): R =  {(0,0),(1,1),(1,2),(1,3),(2,0),(2,2),(2,3), (3,0),(3,3)}

R is the reflexive, because ( a , a )∈R of every elements∈A.

R is the antisymmetric, because if the ( a , b )∈R and if the ( b , a )∈R, then a = b ( since ( 1 . 2 )∈R and ( 2 . 1 )∉R; similarly, all other elements not of the form (a,a) ).

R is not transitive, because ( 1 , 2 )∈R and ( 2 , 0 )∈R, while ( 1 . 0 )∉R.

R is not a partial ordering, because R is not transitive,

e):  R = { ( 0 , 0 ) , ( 0, 1 ) , ( 0 , 2 ) , ( 0 , 3 ) , ( 1 , 0 ) , ( 1 , 1 ) , ( 1 , 2 ) , ( 1 , 3 ) , ( 2 , 0 ) , ( 2 , 2 ) , ( 3 , 3 ) }

R is the reflexive , because ( a , a )∈R of every element a∈A .

R is not antisymmetric, because ( 1 , 0 )∈R and ( 0 , 1 )∈R while 0 is not equal to 1.

R is not transitive, because ( 2 , 0 )∈Rand ( 0 , 3 )∈R, while ( 2 , 3 )∉R .

R is not a partial ordering, because R is not the antisymmetric and not the transitive.

3 0
3 years ago
What is the value of x?
Elenna [48]

Answer:

The answer is 2.

This is because on the left side of the triangle, we can see that the original length is ten cm greater than the second variation. therefore, the other side would also be 10 cm greater than the other variation's side. 5(2)×10= 20, which keeps the triangle to scale.

If this answer was helpful please consider giving brainliest!

5 0
3 years ago
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