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bagirrra123 [75]
3 years ago
8

I am having a hard time on this question. Can you please explain it to me? Please keep it simple so I can understand it. I have

already checked them and I just have this feeling that i'm not correct.
John is going to a gym. He must pay a one time membership fee of $75 and a monthly fee of $30. let m=the number of months john has his membership. write an expression to represent this situation. how much will john pay for 6 months of membership?
Mathematics
1 answer:
Iteru [2.4K]3 years ago
3 0

Answer:

30m + 75

$255

Step-by-step explanation:

Since John  has to pay a monthly fee of 0, you can let m represent the number of months and then add the one time membership fee of $75.

30 multiplied by 6 is 180 and 180 plus 75 is 255.

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Let A = {1, 2, 3, 4, 5} and B = {a, b, c, d}. For each of the following relations fromn A to B, answer these questions: Is it a
Lorico [155]

Answer:

(a) This function is neither one-to-one nor onto.

(b) This function is neither one-to-one nor onto.

(c) This relation is not a function.

(d) The function is onto but not one-to-one.

Step-by-step explanation:

Given information: A = {1, 2, 3, 4, 5} and B = {a, b, c, d}

A relation is a function if and only if there exist a unique output for each input.

One-to-one : A function is one-to-one if every element of the function's codomain is the image of at most one element of its domain.

Onto : A function is onto if for every element y in the codomain Y of f there is at least one element x in the domain X of f such that f(x) = y.

(a)

{(1, c) ,(2, c) ,(3, c) ,(4, c) ,(5, d)}

This relation is a function because all x-value has unique y-value.

The above function it not one-to-one because for more than one input we have same output (c have four domains).

The above function it not onto because all element of B are not have preimage (a and b have no preimage).

This function is neither one-to-one nor onto.

Similarly,

(b)

{(1, a ) ,(2, d ) ,(3, a ) ,(4, c ) }

This relation is a function because all x-value has unique y-value.

Here, a have more than one preimage and b have no preimage.

The function is neither one-to-one nor onto.

(c)

{(1, d ) ,(2, d ) ,(3, a ) ,(4, b ) ,(4, d ) ,(4, c )} .

For x=4 we have for than one value of y.

Therefore this relation is not a function.

(d)

{(1, c ) ,(2, b ) ,(3, a ) ,(4, d ) ,(5, a ) }

This relation is a function because all x-value has unique y-value.

Here, a have two preimage. So, this function is not one-to-one.

All elements of B have preimage. So, this function is onto.

The function is onto but not one-to-one.

5 0
3 years ago
Find the area of the rectangle.
Marianna [84]

Answer:

B

Step-by-step explanation:

= (7x+1) • (8x)

= 56x^2 + 8x

8 0
2 years ago
Find the value of x. 2(x-3)-17=13-3(x+2)
Nonamiya [84]
Solve your equation step-by-step.

2(x−3)−17=13−3(x+2)

Simplify both sides of the equation.

2(x−3)−17=13−3(x+2)

Simplify

2x−23=−3x+7

Add 3x to both sides.

2x−23+3x=−3x+7+3x

5x−23=7

Add 23 to both sides.

5x−23+23=7+23

5x=30

Divide both sides by 5.

5x/5 = 30/5

x=6

5 0
3 years ago
Read 2 more answers
the length of a soccer pitch is 20 m less than twice its width. the area of the pitch is 6000m^2. find its dimensions.
Setler79 [48]
X=width of a coccer pitch
2x-20=length of a soccer pitch
Area (rectangule)=length x width
We suggest this equation:
x(2x-20)=6000
2x²-20x=6000
2x²-20x-6000=0
x²-10x-3000=0
We solve this quadratic equation:

x=[10⁺₋√(100-4*1*-3000)]/2=[10⁺₋√(100+12000)]/2=
=(10⁺₋110)/2
we have two solutions:
x₁=(10-110)/2=-50, invalid solution.
x₂=(10+110)/2=60

x=60
2x-20=2(60)-20=120-20=100

Solution: the length is 100 m, and the width is 60 m.

To check:
Area=100 m*60 m=6000 m²
The twice of width is =2(60 m)=120 m,
20 m less than twice its width is: 120 m-20 m=100 m=the length.


8 0
3 years ago
[Geometry Basics] I'm just checking if this is correct:
Leno4ka [110]
\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ x &,& y~) 
%  (c,d)
&&(~ -9 &,& -1~)
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{ x_2 +  x_1}{2}\quad ,\quad \cfrac{ y_2 +  y_1}{2} \right)
\\\\\\
\left( \cfrac{-9+x}{2}~~,~~\cfrac{-1+y}{2} \right)=\stackrel{midpoint}{(8,14)}\implies 
\begin{cases}
\cfrac{-9+x}{2}=8\\\\
-9+x=16\\
\boxed{x=25}\\
-------\\
\cfrac{-1+y}{2} =14\\\\
-1+y=28\\
\boxed{y=29}
\end{cases}

\bf -------------------------------\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ x &,& y~) 
%  (c,d)
&&(~10 &,& 12~)
\end{array}\qquad
\\\\\\
\left( \cfrac{10+x}{2}~~,~~\cfrac{12+y}{2} \right)=\stackrel{midpoint}{(6,9)}\implies 
\begin{cases}
\cfrac{10+x}{2}=6\\\\
10+x=12\\
\boxed{x=2}\\
-------\\
\cfrac{12+y}{2} =9\\\\
12+y=18\\
\boxed{y=6}
\end{cases}
3 0
3 years ago
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