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3 years ago

5

Sams roller rink costs $3 to get into the rink and $1 for every hour. Brads roller rink costs $5 to get into the rink and $0.50

for every hour. For how many hours would the costs be the same?

1 answer:

PIT_PIT [208]3 years ago

8 0

Answer: 3 hours for sam’s and 2 hours for brad’s

Step-by-step explanation: 3 + 3 = 6 and 5 + 1 = 6

6 = 6 lol

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<img src="https://tex.z-dn.net/?f=Given%20%5C%3A%20cotA%20%3D%20%5Csqrt%7B%5Cdfrac%7B1%7D%7B3%7D%7D" id="TexFormula1" title="Giv

ruslelena [56]

<h3>Given :</h3>

$\tt cotA = \sqrt{ \dfrac{1}{3}}$

$\tt \implies cotA = \dfrac{1}{\sqrt{3}}$

<h3>To Find :</h3>

All other trigonometric ratios, which are :

sinA
cosA
tanA
cosecA
secA

<h3>Solution :</h3>

Let's make a diagram of right angled triangle ABC.

Now, From point A,

AC = Hypotenuse

BC = Perpendicular

AB = Base

$\tt We \: are \: given, \: cotA = \dfrac{1}{\sqrt{3}}$

$\tt We \: know \: that \: cot \theta = \dfrac{base}{perpendicular}$

$\tt \implies \dfrac{base}{perpendicular} = \dfrac{1}{\sqrt{3}}$

$\tt \implies \dfrac{AB}{BC} = \dfrac{1}{\sqrt{3}}$

$\tt \implies AB = 1x \: ; \: BC = \sqrt{3}x \: (x \: is \: positive)$

Now, by Pythagoras' theorem, we have

AC² = AB² + BC²

$\tt \implies AC^{2} = (1x)^{2} + (\sqrt{3}x)^{2}$

$\tt \implies AC^{2} = 1x^{2} + 3x^{2}$

$\tt \implies AC^{2} = 4x^{2}$

$\tt \implies AC = \sqrt{4x^{2}}$

$\tt \implies AC = 2x$

Now,

$\tt sin \theta = \dfrac{perpendicular}{hypotenuse}$

$\tt \implies sinA = \dfrac{BC}{AC}$

$\tt \implies sinA = \dfrac{\sqrt{3}x}{2x}$

$\tt \implies sinA = \dfrac{\sqrt{3}}{2}$

$\Large \boxed{\tt sinA = \dfrac{\sqrt{3}}{2}}$

$\tt cos \theta = \dfrac{base}{hypotenuse}$

$\tt \implies cosA = \dfrac{AB}{AC}$

$\tt \implies cosA = \dfrac{1x}{2x}$

$\tt \implies cosA = \dfrac{1}{2}$

$\Large \boxed{\tt cosA = \dfrac{1}{2}}$

$\tt tan \theta = \dfrac{perpendicular}{base}$

$\tt \implies tanA = \dfrac{BC}{AB}$

$\tt \implies tanA = \dfrac{\sqrt{3}x}{1x}$

$\tt \implies tanA = \sqrt{3}$

$\Large \boxed{\tt tanA = \sqrt{3}}$

$\tt cosec \theta = \dfrac{hypotenuse}{perpendicular}$

$\tt \implies cosecA = \dfrac{AC}{BC}$

$\tt \implies cosecA = \dfrac{2x}{\sqrt{3}x}$

$\tt \implies cosecA = \dfrac{2}{\sqrt{3}}$

$\Large \boxed{\tt cosecA = \dfrac{2}{\sqrt{3}}}$

$\tt sec \theta = \dfrac{hypotenuse}{base}$

$\tt \implies secA = \dfrac{AC}{AB}$

$\tt \implies secA = \dfrac{2x}{1x}$

$\tt \implies secA = 2$

$\Large \boxed{\tt secA = 2}$

5 0

3 years ago

Read 2 more answers

I need help with this question. It is a file.

olchik [2.2K]

D because coefficient is the number next to the variable and 7 is the coefficient to the fifth power

7 0

3 years ago

What is the area of 6in 6in1 2in

Nadya [2.5K]

Answer:

72

Step-by-step explanation:

4 0

3 years ago

Find a number, twice of which decreased by 7 gives 65.<br><br>

damaskus [11]

Answer:

Number = 36

Step-by-step explanation:

Given that,

A number twice of which decreased by 7 gives 65.

Let the number be x.

Twice of the number is 2x.

ATQ,

2x - 7 = 65

We can solve the above equation.

Adding 7 both sides.

2x-7+7 = 65 +7

2x = 72

x = 36

So, the required number is 36.

7 0

3 years ago

A triangle with an area of 2/3 cm? is dilated by a factor of 6. What is the area of the dilated triangle? Enter your answer in t

Y_Kistochka [10]

Answer:

Let A be the area of the dilated triangle.

Given the statement: A triangle with an area of 2/3 cm2 is dilated by a factor of 6.

⇒ Pre-image triangle has an area = $\frac{2}{3}$ $cm^2$

Scale factor(k) states that the dimension of the Dilated-image is 6 times longer than the dimensions of the pre-image.

i.e,

$k = \frac{\text{Dilation image}}{\text{Pre-image}}$

Substitute the given values to find the value of area of dilated triangle.

$6 = \frac{A}{\frac{2}{3} }$

or

$6 = \frac{3A}{2}$

Multiply both sides by 2 we get;

$12 = 3A$

Divide both sides by 3 we get;

$4 = A$

Therefore, the area of the dilated triangle is, $4 cm^2$

5 0

3 years ago