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IRINA_888 [86]
2 years ago
9

The cost of a parking permit consists of a one-time administration fee plus a monthly fee. A permit purchased for 12 months cost

s $660. A permit purchased for 15 months costs $810. What is the administration fee?
Mathematics
2 answers:
LekaFEV [45]2 years ago
5 0
12x50=600 so 60 left
15x50=750 so 60 left
so the administration fee is 60
raketka [301]2 years ago
5 0

Answer:

a=$50 One time administration fee

Step-by-step explanation:

This is a function. Let's model it. Knowing that the Cost of Parking permit is made up of One time administration fee for all, and a monthly payment for everyone.  What's changing here is the amount of Parking Time. So Let's model and solve this Linear System.

C(t)= Cost of Parking Permit. (t= time is given in months)

a= One time administration fee

b = monthly payment

First Case: A permit purchased for 12 months costs $660.

660=12a +b

Second case: A permit purchased for 15 months costs $810

810=15a+b

                                 

660=12a+b    * -5         <em>Addition Method</em>

810=15a+b  * 4

-3300=-60a-5b

 3240= 60a+4b

--------------------------

-60=-b

b=60 monthly fee

                                        Plug in the first equation to find out a.

660 =12a +60

660-60=12a

600=12a

a=$50 One time administration fee

Testing it

$660= 12(50)+60 as 810=15(50)+60 .

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What is 200 percent of (0.020(5/4) + 3 ((1/5) - (1/4)))
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Answer:

0.1/4-3/20=1/40-6/40=-1/8 200% of this is -1/4

Step-by-step explanation:

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2 years ago
Given that r||s and q is a transversal, we know that by the [________]. corresponding angles theorem alternate interior angles t
neonofarm [45]

Answer:

alternate interior angles theorem

Step-by-step explanation:

The alternate interior angles theorem states that when two parallel lines are cut by a transversal, the resulting angles produced are a pair of congruent alternate interior angles.

Given the image attached below, both line k and line l are parallel to each other and also, line t is the transversal, therefore the resulting congruent alternate interior angles produced are:

∠ 4 ≅ ∠6, ∠1 ≅ ∠7

5 0
1 year ago
Read 2 more answers
A lock has 5 buttons. The lock is opened by pushing two buttons simultaneously and then by pushing one button alone. How many co
Pavel [41]

We first must calculate how many ways 2 oblects can be chosen from 5.

combinations = 5! / 2! * (5-2)!

combinations = 5*4 / 2

combinations = 10


There are 10 ways to choose the 2 buttons and 5 ways to choose the final butto so there are 10 * 5 = 50 different ways.


Source

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7 0
1 year ago
Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?
irakobra [83]

First, note that m\angle A+m\angle C=90^{\circ}. Then

m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.

Consider all options:

A.

\tan A=\dfrac{\sin A}{\sin C}

By the definition,

\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.

Now

\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.

Option A is true.

B.

\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.

Then

\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.

Option B is false.

3.

\sin C = \dfrac{\cos A}{\tan C}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

Now

\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.

Option C is false.

D.

\cos A=\tan C.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

As you can see \cos A\neq \tan C and option D is not true.

E.

\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.

Then

\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.

This option is false.

8 0
1 year ago
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