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notka56 [123]
3 years ago
14

Patrick wanted to gather data about the cost of joining a community pool in his area. He plotted the data and determined that th

e average community pool membership cost consists of a one-time registration fee and a monthly fee modeled by the equation y = 15x + 40. Identify and interpret the y-intercept in this model. The y-intercept is 15. This is the cost of registration. The y-intercept is 15. This is the cost per month. The y-intercept is 40. This is the cost per month. The y-intercept is 40. This is the cost of registration.
Mathematics
1 answer:
ohaa [14]3 years ago
4 0

Answer:

The y-intercept is 40. This is the cost of registration ⇒ last answer

Step-by-step explanation:

* Lets explain how to solve the problem

- Patrick wanted to gather data about the cost of joining a community

 pool in his area

- He plotted the data and determined that the average community

 pool membership cost consists of a one-time registration fee and

 a monthly fee

∴ The cost consists of two parts:

# Part A: one-time registration fee, it is a fixed amount not depend on

  any other condition

# Part B: monthly fee, it depends on the number of months

- He modeled that by the equation y = 15 x + 40, where

# y is the cost for the membership

# 15 is the monthly fee

# x is the number of months

# 40 is the one-time registration

- The form of the equation is y = mx + c, where c is the y-intercept

∴ The y-intercept in the equation y = 15 x + 40 is 40

∴ The y-intercept is the one-time registration fee

* The y-intercept is 40. This is the cost of registration

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Step-by-step explanation:

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If y-3x=9, 7x+y=25, what is the value of x and y?
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\bold{\rm{Given}}\text{ : If y - 3x = 9, 7x + y = 25, what is the value of x and y}?

\bold{\rm{Solution}} \text{: We'll solve this by substitution method}

\dashrightarrow  y - 3x = 9 \\  \\  \dashrightarrow   \boxed{y = 9 + 3x } \qquad .. \sf eq(1)

\text{we got the value of y now getting the value of x}

\looparrowright 7x + y = 25 \\  \\  \looparrowright 7x = 25 - y \\  \\  \looparrowright  \boxed{x =   \frac{25 - y}{7}} \qquad .. \sf eq(2)

\text{Now, putting y = 9 + 3x in eq(2)}

\hookrightarrow  x =  \frac{25 - y}{7}   \\  \\ \hookrightarrow  x =  \frac{25 - 9 + 3x}{7}  \\  \\ \hookrightarrow  7x =  25 - 9 + 3x \\  \\ \hookrightarrow  7x - 3x =  16 \\  \\ \hookrightarrow  4x = 16 \\  \\ \hookrightarrow  x =  \frac{16}{4}  \\  \\ \star \quad  \boxed{ \green{ \frak{ x = 4}}}

\text{Now we know the value of x, for getting y we need to put x in eq(1)}

: \implies y = 9 + 3x \\  \\   : \implies y = 9 + 3(4) \\  \\  : \implies y = 9 + 12  \\  \\    \star \quad  \boxed{\frak{  \green{y = 21}}}

\text{Hence, values of x and y are} \:  \frak{ \green{4}}  \: \text{and} \:  \frak{ \green{21}} \: \text{respectively}.

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Answer:

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Step-by-step explanation:

As long as the two figures are congruent, angle P corresponds to angle T, angle [cut off but it says 93 degrees] corresponds to angle U, angle N corresponds to angle V, and angle M corresponds to angle S.

Since we already have the value of T, we know the value of P. The measure of angle P is 61 degrees.

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