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Greeley [361]
2 years ago
5

Two school groups go to a roller skating rink. One group pays $243 for 36 admissions and 21 skate rentals. The other group pays

$81 for 12 admissions and 7 skate rentals. Let x represent the cost of admission and let y represent the cost of a skate rental. Is there enough information to find values for x and y? Explain.
thxs
Mathematics
2 answers:
Masteriza [31]2 years ago
7 0

Answer:

No

There are infinitely many solutions to the system.

{36x+21y=24312x+7y=81

Multiply the second equation by 3.

3(12x+7y36x+21y=81)=243

Subtract this new equation from the first equation.

36x+21y−(36x+21y0+0=243=243)=0 true

Evgen [1.6K]2 years ago
7 0

Answer:

Step-by-step explanation:

this is how to be focused in class

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2. Consider the circle x² + y2 = 1, given in figure. Let OP makes an angle 30° with the x axis.
PolarNik [594]

The equation of the tangent line to the circle passing through the point P is; y = (1/√3)x ± 2/√3

<h3>How to find the equation of the tangent?</h3>

I) We are given the equation of the circle as;

x² + y² = 1

Since angle of inclination is 30°, then slope is;

m = tan 30 = 1/√3

Then equation of the tangent will be;

y = (1/√3)x + c

Put  (√3)x + c into the given circle equation to get;

x² + ((1/√3)x + c)² = 1

x² + ¹/₃x² +  (2/√3)cx + c² = 1

⁴/₃x² +  (2/√3)x + (c² - 1) = 0

Since we need to find value of c for equation to become tangent, then the above quadratic equation must have real and equal roots.

Thus;

((2/√3)c)² - 4(⁴/₃)(c² - 1) = 0

⁴/₃c² - ¹⁶/₃(c² - 1) = 0

⁴/₃c² - ¹⁶/₃c² + ¹⁶/₃ = 0

4c² = ¹⁶/₃

c² = ⁴/₃

c = √⁴/₃

c = ±²/√3

Thus, equation of tangent is;

y = (1/√3)x ± 2/√3

II) Radius from the given equation is 1. Thus, we will use trigonometric ratio to find the x and y intercept;

x-intercept is at y = 0;

0 = (1/√3)x ± 2/√3

-(1/√3)x = ±2/√3

Intercept is positive. Thus;

x = (2/√3)/(1/√3)

x = 1

y - intercept is positive at x = 0;

y = (1/√3)0 ± 2/√3

y = 2/√3

Read more about Equation of tangent at; brainly.com/question/17040970

#SPJ1

6 0
1 year ago
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