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4 years ago

The total cost to buy 12 balloons was $4. The total cost to buy 19 balloons was $6. What was the average cost of the balloons

2 answers:

7 0

Average cost=total number of balloons/total cost=
total number=12+19=31
total cost=4+6=10

so average cost=31/10=3.1

the average cost was $3.10 per ballon

Whitepunk [10]4 years ago

5 0

The average cost of balloons is $3

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Which hyperbola’s asymptote rectangle has the greatest perimeter?

Vitek1552 [10]

Answer:

The greatest perimeter is in (D) ⇒ the perimeter = 60 units

Step-by-step explanation:

* In the hyperbola

- The standard form of the equation of a hyperbola with

center (h , k) and transverse axis parallel to the x-axis is

(x - h)²/a² - (y - k)²/b² = 1

∵ the length of the transverse axis is 2a

∵ the length of the conjugate axis is 2b

∴ The perimeter of asymptote rectangle is 2(2a + 2b)

* Lets check the answers to find the greatest perimeter

A) (x - 4)²/11² - (y + 2)²/3² = 1

* Compare it with the standard form equation

∵ a = 11 ⇒ 2a = 22

∵ b = 3 ⇒ 2b = 6

∴ The perimeter = 2(22 + 6) = 56

B) (x - 2)²/4² - (y + 1)²/10² = 1

* Compare it with the standard form equation

∵ a = 4 ⇒ 2a = 8

∵ b = 10 ⇒ 2b = 20

∴ The perimeter = 2(8 + 20) = 56

C) (x + 5)²/5² - (y - 3)²/9² = 1

* Compare it with the standard form equation

∵ a = 5 ⇒ 2a = 10

∵ b = 9 ⇒ 2b = 18

∴ The perimeter = 2(10 + 18) = 56

D) (x - 7)²/8² - (y - 2)²/7² = 1

* Compare it with the standard form equation

∵ a = 8 ⇒ 2a = 16

∵ b = 7 ⇒ 2b = 14

∴ The perimeter = 2(16 + 14) = 60

* The greatest perimeter is in (D)

4 0

3 years ago

Read 2 more answers

87th term of 12,0,-12

Elan Coil [88]

Answer:

$a_{87} = -1020$

Step-by-step explanation:

Given:

The given AP is.

12,0,-12,......

$n=87$

The first term of an AP $a=12$

Common difference $d=0-12=-12$

We use the the formula for nth term of an AP.

$a_{n} = a+(n-1)d$

for 87th term

$a_{87} = 12+(87-1)(-12)$

$a_{87} = 12+(86)(-12)$

$a_{87} = 12+(-1032)$

$a_{87} = 12-1032$

$a_{87} = -1020$

Therefore, The 87th term of given series is -1020.

5 0

3 years ago

The braking distance, in feet of a car a Travling at v miles per hour is given.

irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

The braking distance at 25 mph, is approximately 63.7 ft.

The braking distance at 55 mph, is approximately 298.35 ft.

The braking distance at 85 mph, is approximately 708.92 ft.

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

$D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.$

At 25 mph, the braking distance is approximately 63.7 ft.

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945 ft.s

$D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.$

At 55 mph, the braking distance is approximately 298.35 ft.

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

$D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.$

At 85 mph, the braking distance is approximately 708.92 ft.

b. The speed of the car when the braking distance is 450 feet is given as follows;

$450 = 2.6 + \dfrac{v^2}{22}$

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ 98.211 ft./s

Learn more here:

brainly.com/question/18591940

8 0

2 years ago

Solve for x.

nikdorinn [45]

Answer:

The answer is 'x>7 or x<1' I believe.

6 0

3 years ago

Question Use the drawing tools to form the correct answer on the provided grid. For the sequence (1,4,7,10,...), the index start

kow [346]

Answer:

hmm let see about dat am coming let me solve it

3 0

3 years ago

Read 2 more answers