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Novay_Z [31]
3 years ago
14

A function with an input of 4 has an output of 10. Which of the following could not be the function equation?

Mathematics
1 answer:
s2008m [1.1K]3 years ago
7 0
For this you'll substitute the input and output.

A) 10 = 4 + 6
      10 = 10

B) 10 = 2(4) + 3
      10 = 11

C) 10 = 14 - 4
      10 = 10

D)10 = 2.5(4)
     10 = 10

The answer choice B is not the equation for the function because 10≠11.

         
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You can use MathPapa it will help you
6 0
3 years ago
Identify the diameter using the following radius: 16 ft.
Firlakuza [10]
B this answer is 4 it’s correct
7 0
3 years ago
determine the diameter to the nearest inch of a large can of tuna that has a volume 6068 inches and a height of 3.3 inches
anyanavicka [17]

Answer:

The diameter of can is<u> 48 inches</u>.

Step-by-step explanation:

Given:

Volume of can = 6068 inches.

And height(h) = 3.3 inches.

Now, to find the diameter.

Diameter = 2 × radius

Let radius be r.

So, by putting the formula of volume we get the radius:

Volume =\pi r^2h

6068=3.14\times r^2\times 3.3 (taking the value of π=3.14)

6068=10.362\times r^2

<em>Dividing both sides by 10.362 we get:</em>

585.60=r^2

<em>Using square root on both sides we get</em>:

24.199=r

<em>Radius = 24.199 inches.</em>

Then, we get the diameter:

Diameter = 2 × radius

Diameter = 2 × 24.199 = 48.398 inches.

Diameter = 48 inches (rounding to nearest, inch as in the place of tenth it is 3 which is less than 5.)

Therefore, the diameter of can is 48 inches.

7 0
3 years ago
Help me out pleaseee ​
spayn [35]

Answer:

Option B is the correct choice.

Step-by-step explanation:

The graph is attached below.

We have to find the x intercept meaning the value of the point onx-axis when y=0

So we will put the value of zero (0) instead of y in our given equation.

So here we have solved it algebraically.

y=\frac{3}{4}x-3

Putting y=0

0=\frac{3}{4}x-3

0=\frac{3x-12}{4}

Multiplying 4 both sides.

0=3x-12

Adding 12both sides.

12=3x

Dividing with 3 both sides.

x=\frac{12}{3}=4

So the x-intercept of the given equation is 4 which can be written as (4,0) in terms of coordinates.

Option B (4,0) is the correct choice.

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
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